# Is there a formula for $\sum_{n=1}^{k} \frac1{n^3}$?

I am searching for the value of $$\sum_{n=k+1}^{\infty} \frac1{n^3} \stackrel{?}{=} \sum_{n = 1}^{\infty} \frac1{n^3} - \sum_{n=1}^{k} \frac1{n^3} = \zeta(3) - \sum_{n=1}^{k} \frac1{n^3}$$

For which I think I need to know the value of $$\sum_{n=1}^{k} \frac1{n^3}$$

Does anyone know of a formula (and a reference if it is complicated)?

• have you already taken a look at this? en.wikipedia.org/wiki/…
– user67133
Apr 14, 2013 at 16:10
• If you're looking to get asymptotics for the tail sum, it is easier to do so directly with Euler-Maclaurin summation, rather than look at the initial sum (which is in some sense the noise rather than the signal). Apr 14, 2013 at 16:16
• @ErickWong. How would I do that? Apr 14, 2013 at 16:20

We have $$S_n = \sum_{k=n+1}^{\infty} \dfrac1{k^3} = \int_{n^+}^{\infty} \dfrac{d \lfloor t \rfloor}{t^3} = \left. \dfrac{\lfloor t \rfloor}{t^3} \right \vert_{t=n^+}^{\infty} + 3 \int_{n^+}^{\infty} \dfrac{\lfloor t\rfloor dt}{t^4} = -\dfrac1{n^2}+3 \int_{n^{+}}^{\infty} \dfrac{t-\{t\}}{t^4} dt$$ Hence, we get that $$S_n = -\dfrac1{n^2} + 3 \int_{n^+}^{\infty} \dfrac{dt}{t^3} - 3 \int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt = \dfrac1{2n^2} - 3 \underbrace{\int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt}_{\mathcal{O}(1/n^3)}$$ You can get better approximations by repeating the above procedure, and this is called as Euler-Maclaurin Summation. To get a higher order approximation, we need to get a good approximation of $$\displaystyle \int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt$$. This is done as follows. $$\int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt = \int_{n^+}^{\infty} \dfrac{1/2}{t^4}dt+ \int_{n^+}^{\infty} \dfrac{\{t\}-1/2}{t^4}dt = \dfrac1{6n^3}+\int_{n^+}^{\infty} \dfrac{B_1(\{t\})dt}{t^4}\\ = \dfrac1{6n^3} + \overbrace{\left.\dfrac{B_2(\{t\})}{2t^4} \right \vert_{t=n^+}^{\infty}}^0 + 4 \underbrace{\int_{n^+}^{\infty} \dfrac{B_2(\{t\})}{t^5} dt}_{\mathcal{O}(1/n^4)}$$ where $$B_n(x)$$ are Bernoulli polynomials of order $$n$$. Hence, a better asymptotic for $$S_n$$ is $$S_n = \dfrac1{2n^2} - \dfrac1{2n^3} + \mathcal{O}(1/n^4)$$ Crank this repeatedly to get higher order estimates.

The Euler MacLaurin for infinite sums, where the integrand is well-behaved (by which I mean all the derivatives of the function vanish as $$x \to \infty$$) is $$\sum_{k=n}^{\infty} f(k) = \int_{n}^{\infty} f(x)dx + \dfrac{f(n)}2 - \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!} f^{(2k-1)'}(n)$$ where $$B_{2k}$$ are the Bernoulli numbers.

In your case, $$f(x) = 1/x^3$$ and $$f^{(2k-1)'}(x) = -\dfrac{(2k+1)!}{2 \cdot x^{2k+2}}$$.

Hence, we get that $$\sum_{k=n}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} + \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$ $$\sum_{k=n+1}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} - \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$

• I really enjoy this answer. I'm going to see if I can get an even better estimate by doing the Euler-Maclaurin again, it would be really cool if we had that $S_n = \frac{1}{2n^2} + \frac{1}{2n^3} + \frac{1}{2n^4} + \cdots$. Apr 14, 2013 at 17:19
• @GeorgeV.Williams No. It won't be since $$\dfrac1{n^2} + \dfrac1{n^3} + \dfrac1{n^4} + \cdots = \dfrac{1/n^2}{1-1/n} = \dfrac1{n(n-1)}$$which is not what we after. Also, I have changed the limits now from $n+1$ to $\infty$ and hence the asymptotic goes as $\dfrac1{2n^2} - \dfrac1{2n^3} \pm$. You might be intereted in looking at the Euler Maclaurin wiki page, to see how the asymptotic actually goes. Euler Maclaurin is very powerful tool...
– user17762
Apr 14, 2013 at 17:23
• @GeorgeV.Williams I have added the full series expansion.
– user17762
Apr 14, 2013 at 17:48
• @user17762 Thanks, I certainly have to look into Euler-Maclaurin more. Apr 14, 2013 at 17:58
• – user17762
Apr 16, 2013 at 21:31

Not to pile on, but there is a neat expression for finite Riemann zetas, based on the following fact:

$$\int_0^{\infty} dt \frac{t^2 e^{-N t}}{e^t-1} = 2 \sum_{k=N+1}^{\infty} \frac{1}{k^3}$$

You can prove this using a geometric sum-type expansion of the denominator and evaluation of the subsequent integrals. Applying integration by parts, you get the following expansion for large $N$:

$$\int_0^{\infty} dt \frac{t^2 e^{-N t}}{e^t-1} = \frac{1}{N^2} - \frac{1}{N^3} + \frac{1}{2 N^4} + O \left ( \frac{1}{N^5} \right )$$

Therefore a good approximation for the sum is

$$\sum_{k=1}^N \frac{1}{k^3} \approx \zeta(3)-\left[\frac{1}{2 N^2} - \frac{1}{2 N^3} + \frac{1}{4 N^4}\right]$$

Even for small $N$, say, $N=5$, the relative error in the the above approximation is vanishingly small, i.e., less than $0.03\%$. For larger $N \sim 1000$, the error is swamped by machine precision. For even smaller errors, one may use Richardson extrapolation as discussed in Bender & Orszag, p. 375.

The simplest estimate, and a surprisingly accurate one, for some $f(n)$ is $$\int_{n - 1/2}^{n + 1/2} \; \; f(x) dx.$$

I edited your question to correct, anyway $$\sum_{k+1}^\infty \frac{1}{n^3} \approx \int_{k + 1/2}^\infty \; \; x^{-3} dx = \frac{2}{(2k+1)^2},$$ and $$\sum_{k}^\infty \frac{1}{n^3} \approx \int_{k - 1/2}^\infty \; \; x^{-3} dx = \frac{2}{(2k-1)^2},$$ this latter value being slightly larger.

If you do some more careful work, you find that the shift by 1/2 gives very attractive small error, because the first derivative function is integrated out over each interval of unit length. So the error is or order $n^{-4},$ with very little effort. Using the Taylor expansion around $x=n,$ I get

$$\int_{n - 1/2}^{n + 1/2} \; \; f(x) dx \approx f(n) + \frac{f''(n)}{12}.$$ Since $f(x) = x^{-3},$ we find $f''(x) = 12 x^{-5},$ and the sum of $n^{-5}$ out to $\infty$ is approximately a constant times $n^{-4}.$

=-=-=-=-=-=-=-=

const double zeta_3 =    1.20205690315959428539  ;

for (int n = 1; n <= bound; n += 1)
{
sump += 1.0 / (n * n * n);
double diff = zeta_3 - sump;
double est = 2.0 / ( (2.0 * n + 1.0 ) * (2.0 * n + 1.0 ) );
cout.precision(16);
cout << n << "  " << sump << "  " << diff << "  " << est << "  " <<   log(est - diff) / log(1.0 * n)  << endl;

} // for n

n   sump                 diff                 estimate             log ratio
1  1                  0.2020569031595942   0.2222222222222222    -inf
2  1.125              0.07705690315959424  0.08                 -8.408449270798803
3  1.162037037037037  0.04001986612255726  0.04081632653061224  -6.494860116575991
4  1.177662037037037  0.02439486612255726  0.02469135802469136  -5.859859838755892
5  1.185662037037037  0.01639486612255725  0.01652892561983471  -5.540584559448334
6  1.190291666666667  0.01176523649292771  0.01183431952662722  -5.346812200208313
7  1.19320711856171  0.008849784597883881  0.008888888888888889  -5.215697321661032
8  1.19516024356171  0.006896659597883881  0.006920415224913495  -5.12045706241989
9  1.196531985674193  0.005524917485401071  0.00554016620498615  -5.047738467187578
10  1.197531985674193  0.004524917485401181  0.00453514739229025  -4.990128319146853
11  1.198283300475095  0.003773602684499666  0.003780718336483932  -4.9431740534746
12  1.198862004178798  0.003194898980795946  0.0032                -4.904035407248641
13  1.199317170314438  0.002739732845156384  0.002743484224965706  -4.870812283354566
14  1.199681601801318  0.002375301358275905  0.002378121284185493  -4.842183533928056
15  1.199977898097615  0.002079005061979666  0.002081165452653486  -4.817200759352243
16  1.200222038722615  0.001834864436979666  0.001836547291092746  -4.795164613680947
17  1.200425580346877  0.001631322812717606  0.00163265306122449  -4.775547458961447
18  1.200597048110937  0.00145985504865731  0.001460920379839299  -4.757943264762034
19  1.200742841958437  0.001314061201157735  0.001314924391847469  -4.742034155780267
20  1.200867841958436  0.001189061201157804  0.001189767995240928  -4.727567477547702
21  1.200975821658253  0.001081081501341341  0.001081665765278529  -4.714339704360496
22  1.201069736008366  0.0009871671512287072  0.0009876543209876543  -4.70218491428755
23  1.201151925537419  0.0009049776221747852  0.0009053870529651426  -4.690966382055557
24  1.201224263500382  0.0008326396592117646  0.0008329862557267805  -4.680570345424018
25  1.201288263500383  0.0007686396592117006  0.0007689350249903883  -4.670901314294367
26  1.201345159267337  0.0007117438922568109  0.000711997152011392  -4.661878492460928
27  1.201395964530763  0.0006609386288314312  0.0006611570247933885  -4.653433014433951
28  1.201441518466623  0.0006153846929712881  0.0006155740227762388  -4.645505789221971
29  1.20148252055773  0.0005743826018647091  0.0005745475438092502  -4.638045798396245
30  1.201519557594767  0.0005373455648276515  0.0005374899220639613  -4.631008740248393
31  1.201553124779487  0.0005037783801074003  0.0005039052658100278  -4.624355942352548
32  1.201583642357612  0.0004732608019824003  0.0004733727810650888  -4.61805347945706
33  1.201611468831719  0.0004454343278750272  0.0004455335263978615  -4.612071454176108
34  1.201636911534752  0.0004199916248421864  0.0004200798151648813  -4.606383404809224
35  1.201660235149912  0.0003966680096818553  0.0003967466772465781  -4.600965818754339
36  1.20168166862042  0.0003752345391743184  0.0003753049352598987  -4.595797723054888
37  1.201701410787715  0.0003554923718791514  0.0003555555555555556  -4.590860344914228
38  1.201719635018653  0.0003372681409417044  0.000337325012649688  -4.586136825474815
39  1.201736493023676  0.0003204101359179923  0.0003204614645088928  -4.581611975945853
40  1.201752118023676  0.0003047851359179177  0.0003048315805517452  -4.577272072501593
41  1.201766627389472  0.0002902757701221947  0.0002903178980984177  -4.573104680223196
42  1.201780124851949  0.0002767783076451646  0.0002768166089965398  -4.569098502954018
43  1.201792702360848  0.0002642007987465611  0.0002642356982428326  -4.56524325033338
44  1.201804441654612  0.0002524615049823709  0.0002524933720489837  -4.561529528093542
45  1.201815415591512  0.0002414875680825102  0.0002415167250332085  -4.557948742050555
46  1.201825689282644  0.0002312138769506866  0.0002312406058503873  -4.554493008171014
47  1.201835321059803  0.0002215820997915063  0.000221606648199446  -4.551155084738899
48  1.201844363305173  0.0002125398544210455  0.0002125624402168137  -4.547928300412313
49  1.201852863164925  0.000204039994668781  0.0002040608101214162  -4.544806505967626
50  1.201860863164925  0.000196039994668773  0.0001960592098813842  -4.541784013918732
51  1.201868401743602  0.0001885014159923593  0.0001885191818267509  -4.538855561589176
52  1.201875513714471  0.0001813894451230258  0.00018140589569161  -4.53601627043618
53  1.201882230668736  0.0001746724908586739  0.0001746877456546423  -4.533261605315103
54  1.201888581326664  0.0001683218329304736  0.000168335998653312  -4.530587353600314
55  1.201894591845071  0.0001623113145232669  0.000162324486648811  -4.52798958873942
56  1.201900286087054  0.0001566170725406657  0.0001566293366747592  -4.525464645351152
57  1.201905685859183  0.0001512173004110107  0.0001512287334593573  -4.5230091053239
58  1.201910811120572  0.000146092039022605  0.0001461027102052743  -4.520619767225281
59  1.201915680167553  0.0001412229920412233  0.0001412329637737448  -4.518293638760038
60  1.201920309797183  0.0001365933624115634  0.0001366026910730141  -4.516027905959973
61  1.201924715452282  0.0001321877073126032  0.0001321964439156587  -4.513819932780932
62  1.201928911350372  0.0001279918092225163  0.000128  -4.511667243064355
63  1.201932910598513  0.0001239925610811987  0.000124000248000496  -4.50956750844126
64  1.201936725295779  0.0001201778638155737  0.0001201850850309477  -4.507518530365488
65  1.201940366624864  0.000116536534730427  0.0001165433249810617  -4.505518239614971
66  1.201943844934127  0.0001130582254669221  0.0001130646164282888  -4.503564683263746
67  1.20194716981119  0.0001097333484043617  0.0001097393689986283  -4.501656025894788
68  1.201950350149069  0.0001065530105253121  0.0001065586871969737  -4.499790515309039
69  1.201953394205701  0.0001035089538936607  0.0001035143108534755  -4.497966501510966
70  1.201956309657596  0.000100593501998647  0.0001005985614405714  -4.496182427673079
71  1.20195910364828  9.779951131383413e-05  9.780429360848941e-05  -4.49443681084581
72  1.201961782832094  9.512032750036425e-05  9.512485136741974e-05  -4.492728244598787
73  1.201964353413842  9.254974575201302e-05  9.255402841408673e-05  -4.491055399816985
74  1.201966821184754  9.008197484017266e-05  9.008603216071348e-05  -4.489417011148479
75  1.201969191555124  8.771160446974235e-05  8.771545107670716e-05  -4.487811865496496
76  1.201971469583992  8.543357560264475e-05  8.543722499893203e-05  -4.486238832357404
77  1.201973660006152  8.324315344254707e-05  8.324661810613944e-05  -4.484696804868141
78  1.20197576725678  8.113590281455529e-05  8.113919428780072e-05  -4.483184751247111
79  1.201977795493897  7.910766569740879e-05  7.911079466793244e-05  -4.481701682655552
80  1.201979748618897  7.715454069745498e-05  7.715751707110065e-05  -4.480246655628918
81  1.20198163029532  7.527286427433388e-05  7.527569724114569e-05  -4.478818763916544
82  1.201983443966044  7.345919354984076e-05  7.346189164370983e-05  -4.477417139728746
83  1.201985192869045  7.171029054919842e-05  7.171286170174621e-05  -4.476040964568041
84  1.201986880051855  7.002310773951415e-05  7.002555932915514e-05  -4.474689451142798
85  1.201988508384849  6.839477474551714e-05  6.839711364180432e-05  -4.47336186065972
86  1.201990080573461  6.682258613310843e-05  6.682481873767917e-05  -4.47205743944392
87  1.201991599169428  6.530399016613231e-05  6.530612244897959e-05  -4.470775515614187
88  1.201993066581149  6.383657844555302e-05  6.383861597880558e-05  -4.469515423981185
89  1.201994485083239  6.241807635531949e-05  6.24200243438095e-05  -4.468276529506308
90  1.201995856825351  6.104633424275363e-05  6.104819755196727e-05  -4.46705822147225
91  1.201997183840324  5.971931926995744e-05  5.972110245155126e-05  -4.465859915788036
92  1.201998468051716  5.843510787850725e-05  5.843681519357195e-05  -4.464681064582156
93  1.201999711280779  5.719187881481069e-05  5.719351425548342e-05  -4.463521115171593
94  1.202000915252924  5.598790666994091e-05  5.598947397889197e-05  -4.462379551348827
95  1.202002081603704  5.482155588998339e-05  5.482305857843809e-05  -4.46125587352416
96  1.202003211884376  5.369127521870354e-05  5.369271658299552e-05  -4.460149597033682
97  1.202004307567057  5.259559253723012e-05  5.259697567389875e-05  -4.45906026950906
98  1.202005370049526  5.153311006811379e-05  5.153443788811873e-05  -4.457987418363708
99  1.202006400659678  5.050249991600531e-05  5.0503775157193e-05  -4.456930651941094
100  1.202007400659678  4.950249991608757e-05  4.950372515531794e-05  -4.455889553084795


=-=-=-=-=-=-=-=

EEEDDDDIIITTTT:

Do the same trick one more time, $$\sum_{k = 1}^n \frac{1}{k^3} +\frac{2}{(2n+1)^2} - \frac{2}{(2n+1)^4} = \zeta(3) - O(\frac{1}{n^6})$$

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

   est = 2 / (2 n + 1)^2   -   2 / (2 n + 1)^4

const double zeta_3 =    1.20205690315959428539  ;

n  sump                  est                    sump + est
1  1                     0.1975308641975309  1.197530864197531
2  1.125                0.07680000000000001  1.2018
3  1.162037037037037    0.03998334027488546  1.202020377311922
4  1.177662037037037    0.02438652644413961  1.202048563481177
5  1.185662037037037    0.0163923229287617   1.202054359965799
6  1.190291666666667    0.01176429396729806  1.202055960633965
7  1.19320711856171    0.008849382716049382  1.20205650127776
8  1.19516024356171    0.006896469151470888  1.202056712713181
9  1.196531985674193   0.005524819484196714  1.20205680515839
10  1.197531985674193  0.004524863611355351  1.202056849285548
11  1.198283300475095  0.003773571420914019  1.202056871896009
12  1.198862004178798  0.00319488            1.202056884178798
13  1.199317170314438  0.002739720872119389  1.202056891186557
14  1.199681601801318  0.002375293553764345  1.202056895355083
15  1.199977898097615  0.002078999827832827  1.202056897925447
16  1.200222038722615  0.001834860838116536  1.202056899560731
17  1.200425580346877  0.001631320283215327  1.202056900630092
18  1.200597048110937  0.001459853235661184  1.202056901346598
19  1.200742841958437  0.001314059878769331  1.202056901837206
20  1.200867841958436  0.001189060221299678  1.202056902179736
21  1.200975821658253  0.001081080764864641  1.202056902423118
22  1.201069736008366  0.0009871665904587715  1.202056902598824
23  1.201151925537419  0.0009049771901073041  1.202056902727527
24  1.201224263500382  0.0008326393226756656  1.202056902823058
25  1.201288263500383  0.0007686393944540598  1.202056902894837
26  1.201345159267337  0.0007117436820391559  1.202056902949377
27  1.201395964530763  0.0006609384604876717  1.20205690299125
28  1.201441518466623  0.0006153845570874804  1.20205690302371
29  1.20148252055773   0.0005743824913692016  1.202056903049099
30  1.201519557594767  0.0005373454743558011  1.202056903069122
31  1.201553124779487  0.0005037783055515722  1.202056903085038
32  1.201583642357612  0.0004732607401701621  1.202056903097782
33  1.201611468831719  0.0004454342763362892  1.202056903108055
34  1.201636911534752  0.0004199915816393268  1.202056903116391
35  1.201660235149912  0.0003966679732836249  1.202056903123196
36  1.20168166862042   0.0003752345083626834  1.202056903128783
37  1.201701410787715  0.0003554923456790124  1.202056903133394
38  1.201719635018653  0.0003372681185676084  1.20205690313722
39  1.201736493023676  0.0003204101167337752  1.20205690314041
40  1.201752118023676  0.0003047851194054944  1.202056903143082
41  1.201766627389472  0.0002902757558574396  1.202056903145329
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n    sump                  est                    sump + est

const double zeta_3 =    1.20205690315959428539  ;

est = 2 / (2 n + 1)^2   -   2 / (2 n + 1)^4


=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

For any decreasing $f(x)$, we have that:

$$\int_{a}^{b+1} f(i) \; \mathrm d i \le \sum_{i=a}^b f(i) \le \int_{a-1}^b f(i) \; \mathrm d i$$

So in our case, with $b = \infty$, and $a = k$:

$$\int_{k}^{\infty} i^{-3} \; \mathrm d i \le \sum_{i=k}^\infty i^{-3} \le \int_{k-1}^\infty i^{-3} \; \mathrm d i$$

Giving:

$$\int_{k}^{\infty} i^{-3} \; \mathrm d i \le \sum_{i=k}^\infty i^{-3} \le \int_{k-1}^\infty i^{-3} \; \mathrm d i$$

$$\frac{1}{2k^2} \le \sum_{i=k}^\infty i^{-3} \le \frac{1}{2(k-1)^2}$$

• is there a theorem which states that the first identity is true? Apr 14, 2013 at 17:11
• @user27182, I'm not sure if it is named, but consider this intuitive proof: The sum is the sum of rectangles from $a$ to $b-1$ with length $1$. The integrals are the area under the curve. If the function is decreasing, then the rectangle bounds the curve. Apr 14, 2013 at 17:12
• @GeorgeVWilliams. Yes, I saw that argument, I was just wondering if it is dependent on some fact about Reimann sums or something similar. I've never come across any type off sum-integral inequalities in my analysis courses. Apr 14, 2013 at 17:20

$$\sum _{n=1}^k \frac{1}{n^s}=H_k^{(s)}$$ Where $H_k^{(s)}$ - is the harmonic number of order s.

A related problem. No need for the calculations you are doing. Just recall the Herwitz zeta function

$$\zeta(s,a)=\sum_{n=0}^{\infty}\frac{1}{(n+a)^3},$$

then you can readily write the sum $\sum_{n=k+1}^{\infty} \frac1{n^3}$ in terms of it as

$$\sum_{n=k+1}^{\infty} \frac1{n^3} = \sum_{n=0}^{\infty} \frac1{(n+k+1)^3}=\zeta(3,k+1).$$

For asymptotic behavior of the Herwitz zeta function with respect to $a$ and $s$, see here.

• great knowledge! Apr 14, 2013 at 19:16
• @user27182: You are welcome. Apr 14, 2013 at 19:17

Suppose we answer this question by computing the asymptotics of $$\sum_{k=1}^n \frac{1}{k^3}.$$ This is actually a textbook example of harmonic summation techniques.

Introduce $$S(x) = \sum_{k\ge 1} \left(\frac{1}{k^3}-\frac{1}{(x+k)^3}\right)$$ so that our answer for the tail end of the series is given by $$\zeta(3)-S(n).$$

Re-write $S(x)$ as follows: $$S(x) = \sum_{k\ge 1} \frac{1}{k^3} \left(1-\frac{1}{(x/k+1)^3}\right).$$ The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^3}, \quad \mu_k = \frac{1}{k} \quad \text{and} \quad g(x) = 1-\frac{1}{(x+1)^3}.$$

Canceling an initial part of the series about zero of the function being transformed only shifts the fundamental strip, so we may start by calculating the Mellin transform $h^*(s)$ of $h(x)=1-g(x)$ which is $$\int_0^\infty \frac{1}{(1+x)^3} x^{s-1} dx.$$

Now this Mellin transform can be evaluated using a keyhole contour with the slot on the positive real axis, which gives

$$(1-e^{2\pi i (s-1)})\times \int_0^\infty \frac{1}{(1+x)^3} x^{s-1} dx = 2\pi i \times \mathrm{Res}\left(\frac{1}{(1+x)^3} x^{s-1}; x= -1\right).$$

The branch of the logarithm used with this contour has the cut on the positive real axis so that in a neighborhood of $x=-1$ using the binomial series $$x^{s-1} = \sum_{q\ge 0} \left.\left(\frac{d}{dx}\right)^q x^{s-1}\right|_{x=-1} \frac{(x+1)^q}{q!}$$ and the residue is $$\frac{1}{2!} \left.\left(\frac{d}{dx}\right)^2 x^{s-1}\right|_{x=-1} \\= \frac{1}{2} (s-2) (s-1) (-1)^{s-3} = \frac{1}{2} (s-2) (s-1) e^{i\pi(s-3)} = \frac{1}{2} (s-2) (s-1) e^{i\pi(s-1)}$$ and we have for the transform that $$h^*(s) = \int_0^\infty \frac{1}{(1+x)^3} x^{s-1} dx = 2\pi i \frac{1}{2} (s-1) (s-2) \frac{ e^{\pi i (s-1)}}{1-e^{2\pi i s}} \\= -\frac{1}{2} (s-1) (s-2) 2\pi i \frac{1}{e^{-\pi i s}-e^{\pi i s}} = \frac{1}{2} (s-1) (s-2) \frac{\pi}{\sin(\pi s)}.$$

It follows that the Mellin transform of $g(x)$ is also given by $$g^*(s) = -\frac{1}{2} (s-1) (s-2)\frac{\pi}{\sin(\pi s)}.$$

The Mellin integral that we just computed has fundamental strip $\langle 0,3\rangle$. Canceling the first term of the expansion about zero by adding one keeps the transform but shifts the fundamental strip to $\langle -1,0\rangle.$

Therefore the transform $Q(s)$ of $S(x)$ is $$Q(s) = -\frac{1}{2} (s-1) (s-2) \frac{\pi}{\sin(\pi s)} \zeta(3-s) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1}\frac{1}{k^3} k^s = \zeta(3-s).$$ The half plane of convergence of the zeta term is $\Re(s)<2.$

Taking into account the intersection of $\langle -1, 0\rangle$ with the half plane of convergence we thus obtain the Mellin inversion integral

$$S(x) = \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the right for an expansion about infinity.

We get $$\mathrm{Res}(Q(s)/x^s; s = 0) = -\frac{1}{2} \times -1 \times -2 \times (-1)^0 \times \zeta(3) = -\zeta(3).$$

(Not to worry about the apparent sign error, since the residues will be negated because we are shifting to the right.)

The factor $(s-1)$ cancels the pole at $s=1$ from the sine term.

The factor $(s-2)$ cancels the pole from the sine term but there is a contribution from the zeta function term which gives $$\mathrm{Res}(Q(s)/x^s; s = 2) = -\frac{1}{2} \times 1 \times (-1)^2 \times -1 \times \frac{1}{x^2} = \frac{1}{2} \frac{1}{x^2}.$$

The pole at $s=3$ is the last one to merit special treatment. We get $$\mathrm{Res}(Q(s)/x^s; s = 3) = -\frac{1}{2} \times 2 \times 1 \times (-1)^3 \times \zeta(0) \times \frac{1}{x^3} \\= \frac{1}{4} \times -2 \times \frac{1}{x^3} = - \frac{1}{2} \frac{1}{x^3}.$$

For the remaining poles at $s=q$ where $q>3$ we obtain $$\mathrm{Res}(Q(s)/x^s; s = q) = -\frac{1}{2} (q-1) (q-2) (-1)^q \zeta(3-q)\frac{1}{x^q} \\= -\frac{1}{2} (-1)^q (q-1) (q-2) \zeta(-(q-3))\frac{1}{x^q} = \frac{1}{2} (-1)^q (q-1) (q-2) \frac{B_{q-2}}{q-2} \frac{1}{x^q} = \frac{1}{2} (-1)^q (q-1) B_{q-2}\frac{1}{x^q} .$$ Now this only contributes when $q$ is even so that we may simplify it to $$\frac{1}{2} \sum_{q\ge 2} (2q-1) B_{2q-2} \frac{1}{x^{2q}}.$$

This gives for the asymptotic expansion $$S(x) \sim \zeta(3) - \frac{1}{2x^2} + \frac{1}{2x^3} - \frac{1}{2} \sum_{q\ge 2} (2q-1) B_{2q-2} \frac{1}{x^{2q}}.$$

We finally have $$\zeta(3) - S(n) \sim \frac{1}{2n^2} - \frac{1}{2n^3}+ 1/4\,{n}^{-4}-1/12\,{n}^{-6}+1/12\,{n}^{-8}-{\frac {3}{20}}\,{n}^{-10} \\+{\frac {5}{12}}\,{n}^{-12}-{\frac {691}{420}}\,{n}^{-14}+{\frac {35}{ 4}}\,{n}^{-16}-{\frac {3617}{60}}\,{n}^{-18}+{\frac {43867}{84}}\,{n}^ {-20}\\-{\frac {1222277}{220}}\,{n}^{-22}+{\frac {854513}{12}}\,{n}^{-24 }-\cdots$$

Consult this MSE link for a closely related computation.

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\on}{\operatorname{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ Indeed, there is a Zeta Function related identity: \begin{align} &\bbox[5px,#ffd]{\sum_{n = k + 1}^{\infty} {1 \over n^{3}}} = \zeta\pars{3} - \sum_{n = 1}^{k}{1 \over n^{3}} \\[5mm] = &\ \zeta\pars{3} - \underbrace{\bracks{% \zeta\pars{\color{red}{3}} - {k^{1 - \color{red}{3}} \over \color{red}{3} - 1} + \color{red}{3}\int_{k}^{\infty}{x - \left\lfloor x\right\rfloor \over x^{\color{red}{3} + 1}}\dd x}}_{\ds{\sum_{n = 1}^{k}{1 \over n^{\color{red}{3}}}}}\ \\[5mm] = &\ \bbx{{1 \over 2k^{2}} - 3\int_{k}^{\infty}{x - \left\lfloor x\right\rfloor \over x^{4}}\dd x} \\[5mm] = & {1 \over 2k^{2}} - {1 \over 2k^{3}} + {1 \over 4k^{4}} + \mrm{O}\pars{1 \over k^{5}} \end{align}