# Contest problem about convergent series

The following is probably a math contest problem. I have been unable to locate the original source.

Suppose that $\{a_i\}$ is a set of positive real numbers and the series $$\sum_{n = 1}^\infty \frac{1}{a_n}$$ converges.

Show $$\sum_{n = 1}^\infty \frac{n^2a_n}{(a_1+\cdots+a_n)^2}$$

also converges.

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Just to record what transpired below - one line of attack that doesn't seem to work is to try to bound the arithmetic mean by the harmonic mean. If $A_n = (a_1 + \ldots + a_n)/n$ and $H_n = n(1/a_1 + \ldots + 1/a_n)^{-1}$, it's well known that $A_n \ge H_n$, and we might desire to bound the series as $\sum_n a_n/A_n^2 \le \sum_n a_n/H_n^2$. However, such a bound is too weak because $1/H_n$ goes like $1/n$, so if we pick a counterexample like $a_n = n^3$, which satisfies the hypotheses, then this upper bound would actually diverge. – Christopher A. Wong Sep 22 '12 at 9:49
I am almost certain this was a potw for Purdue. But I don't have the time to search through all of them. Maybe later. – user641 Sep 22 '12 at 16:38
POTW ... President of the World – GEdgar Sep 22 '12 at 17:55
@Steve D In the Purdue Problems, the Fall 2006 problem 6 is similar. The part of the solution that upgrades from increasing $\{a_n\}$ to general $\{a_n\}$ seems like it would still work. But I've been unsuccessful in my attempts to adapt the first part of the given solution. – alex.jordan Sep 23 '12 at 10:39
I just came across this problem. I think its original source is the contest william putam. It was the third problem of the second day year 1966. Here is a url with the problems and solutions math-olympiad.com/… – clark Sep 29 '12 at 21:54

Define at first some quantities to simplify the typing for the rest of the proof

• $$C^2:=\sum_{n=1}^{+\infty}\frac{1}{a_n}.$$
• $$A_n=a_1+\dotso+a_n.$$

Moreover let $$P_N:=\sum_{n=1}^N\frac{n^2a_n}{(a_1+\dotso+a_n)^2}.$$ Clearly $P_{N+1}>P_N$, that is, $P_N$ is an increasing sequence. If we can prove that it is also bounded above, we are done with the proof. To reach this goal, notice that $$\begin{split}P_N<&\frac{1}{a_1}+\sum_{n=2}^N\frac{n^2(A_n-A_{n-1})}{A_nA_{n-1}}\\ =&\frac{1}{a_1}+\sum_{n=2}^N\left(\frac{n^2}{A_{n-1}}-\frac{n^2}{A_n}\right).\end{split}\tag{1}$$ Since $(n+1)^2-n^2=2n+1<5n$ for every $n\in\mathbb N$, one gets from $(1)$ that $$\begin{split}P_N<&\frac{1}{a_1}+\frac{4}{a_1}+\sum_{n=2}^{N-1}\frac{2n+1}{A_n}-\frac{N^2}{A_N}\\ <&\frac{5}{a_1}+\frac{5}{A_2}+\dots+\frac{2N-1}{A_{N-1}}-\frac{N^2}{A_N}\\<&5\left(\frac{1}{A_1}+\frac{2}{A_2}+\dots+\frac{N}{A_N}\right).\end{split}\tag{2}$$ By Cauchy Schwarz we also have $$\sqrt{\left(\frac{1}{a_1}+\dots+\frac{1}{a_N}\right)}\sqrt{\left(\frac{a_1}{A_1^2}+\dots+\frac{N^2a_N}{A_N^2}\right)}\geq\left(\frac{1}{A_1}+\frac{2}{A_2}+\dots+\frac{N}{A_N}\right),\tag{3}$$ from which, following $(2)$, it turns out that $$P_N<5C\sqrt{P_N}.$$ It is then clear that the sequence $P_N$ is bounded from above, since for any $N\in\mathbb N$, we have estabilished $$P_N<25C^2.$$ Therefore, since $P_N$ is also increasing as observed at the beginning, we can conclude that $P_N$ converges. This concludes the proof.

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Fantastic solution, thank you. – Potato Sep 25 '12 at 0:57
Note that the right side of (2) is bounded above by the solution to the Purdue problem linked to in the comments to the original question. – Zarrax Sep 25 '12 at 1:03
+1. details i needed to think about, for $n \geq 2, \; a_n = A_n - A_{n-1}.$ Next $\frac{1}{A_n} < \frac{1}{A_{n-1}}.$ Finally the Cauchy-Schwarz step has one vector $x_n = \frac{1}{\sqrt a_n},$ the other $y_n = \frac{n \sqrt a_n}{A_n}.$ – Will Jagy Sep 25 '12 at 2:17
Inequality (2) reduces the question to the Purdue problem, and (3) performs the reverse reduction from Purdue to this question. – zyx Sep 25 '12 at 9:54
Well done. Bravo ! – Did Sep 25 '12 at 12:37

I wrote this answer for the closed duplicate of this question, but it works here as well.

Define $$\bar{p}_n=\frac1n\sum_{k=1}^np_k\tag{1}$$ then the series in question is $$\sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\tag{2}$$ Simply, for $n\ge m$, we have that $$\bar{p}_n=\frac1n\sum_{k=1}^np_k\ge\frac mn\frac1m\sum_{k=1}^mp_k=\frac mn\bar{p}_m\tag{3}$$ which, for $n\ge1$, says that $$\bar{p}_{n+1}\ge\frac12\bar{p}_n\quad\text{and}\quad\bar{p}_{2n+1}\ge\frac23\bar{p}_{2n}\tag{4}$$ Furthermore, \begin{align} \sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}\bar{p}_k} &=\sum_{k=1}^\infty\frac{(k+1)\bar{p}_{k+1}-k\bar{p}_k}{\bar{p}_{k+1}\bar{p}_k}\\ &=\sum_{k=1}^\infty\left(\frac{k}{\bar{p}_k}+\frac1{\bar{p}_k}-\frac{k+1}{\bar{p}_{k+1}}+\frac1{\bar{p}_{k+1}}\right)\\ &=2\sum_{k=1}^\infty\frac1{\bar{p}_k}\tag{5} \end{align} Combining $(4)$ and $(5)$ yields \begin{align} \sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2} &=\frac1{p_1}+\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}^2}\\ &\le\frac1{p_1}+2\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}\bar{p}_k}\\ &=\frac1{p_1}+4\sum_{k=1}^\infty\frac1{\bar{p}_k}\tag{6} \end{align} Use $\color{#C00000}{(4)}$, $\color{#00A000}{\text{Jensen's Inequality}}$, and change the $\color{#0000FF}{\text{order of summation}}$ to get \begin{align} \sum_{k=1}^\infty\frac1{\bar{p}_k} &=\frac1{p_1}+\sum_{k=1}^\infty\left(\frac1{\bar{p}_{2k}}+\frac1{\bar{p}_{2k+1}}\right)\\ &\le\frac1{p_1}+\color{#C00000}{\frac52\sum_{k=1}^\infty\frac1{\bar{p}_{2k}}}\\ &\le\frac1{p_1}+5\sum_{k=1}^\infty\frac1{\displaystyle\small\frac2{2k}\sum_{j=k+1}^{2k}p_j}\\ &\le\frac1{p_1}+5\sum_{k=1}^\infty\color{#00A000}{\frac1k\sum_{j=k+1}^{2k}\frac1{p_j}}\\ &=\frac1{p_1}+5\color{#0000FF}{\sum_{j=2}^\infty\frac1{p_j}\sum_{k=\lceil j/2\rceil}^{j-1}\frac1k}\\ &\le\frac1{p_1}+5\sum_{j=2}^\infty\frac1{p_j}\tag{7} \end{align} Combining $(6)$ and $(7)$ gives $$\sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\le20\sum_{j=1}^\infty\frac1{p_j}\tag{8}$$

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I printed out the Purdue problem. I also felt it worthwhile to do some experiments. If we stick with $a_n = n^k$ for fixed $k \geq 2,$ things are pretty good. So what I did was run $$a_n = (n + 14) \log(n+14) \left( \log \log (n+14) \right)^2 \; \; / \; \; 15.$$ Note that, in contrast to the Purdue problem it is possible for the ratio to exceed $e.$ Also, the ratio is steadily increasing, with the appearance of approaching a limit. I have the feeling that if I changed the "fancy" sum to have $(n + 14)^2$ instead of $n^2$ in the numerator, the ratio might easily stay below $e.$ I should try that. NO: with the $(n + 14)^2$ the ratio starts at 225 and decreases. Annoying, but better to know the truth.

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    n   harmonic sum           fancy sum             ratio
1   0.3720703157450488    0.3720703157450488      1
2   0.697211225420433    0.7424549077736787      1.064892360741844
3   0.9843674932090888    1.109100676396848      1.126714041298867
4   1.240296228954154    1.47046914150556      1.185578982809206
5   1.470203813539751    1.825431331069707      1.241617872473538
6   1.678171381303085    2.173181689415959      1.29497005706801
7   1.867446704822137    2.513168481471731      1.345777887520006
8   2.040649149703482    2.845038024641238      1.394182838855292
9   2.199916583903517    3.168590269249502      1.440322916074925
10   2.347012629961203    3.483743567272218      1.484330984332967
11   2.483406263112077    3.790506814518016      1.526333758121375
12   2.610331767584863    4.088957475567311      1.566451255868715
13   2.728834508384516    4.379224284501249      1.604796579288999
14   2.839806304031565    4.661473653263476      1.641475915679798
15   2.94401306997052    4.935899015735965      1.676588689800005
16   3.042116644209738    5.202712494418531      1.71022781270442
17   3.134692183052954    5.462138403848396      1.742479990022078
18   3.222242147523305    5.714408206199048      1.773426063150246
19   3.305207639920737    5.959756614818636      1.803141364807435
20   3.383977661836249    6.198418604998871      1.831696076160095
21   3.458896727815719    6.430627141452748      1.859155576903756
22   3.530271167787264    6.656611471609346      1.885580782674448
23   3.598374376086184    6.876595865132866      1.911028466307689
24   3.663451208300954    7.090798704807332      1.935551560981734
25   3.72572168420513    7.299431853492455      1.959199444348663
26   3.785384122162883    7.502700237347626      1.982018203494961
27   3.842617805032313    7.700801597802712      2.004050881073236
28   3.897585257875515    7.893926374503574      2.025337703274867
29   3.950434202350573    8.082257689209232      2.045916290518181
30   4.001299240496784    8.26597140678431      2.065821851843813
31   4.050303310976641    8.445236254343797      2.085087364063956
32   4.097558953139548    8.620213983526074      2.103743736724342
33   4.143169408093855    8.791059563998846      2.12181996392065
34   4.187229580988316    8.957921398802352      2.139343263974555
35   4.229826884660201    9.120941554132628      2.156339207926083
36   4.271041981511019    9.280255997765739      2.172831837743385
37   4.310949437771172    9.43599484160206      2.188843775092063
38   4.349618302093892    9.5882825848318      2.204396321446421
39   4.387112618583768    9.737238355039727      2.219509550266131
40   4.423491882842733    9.882976145219228      2.234202391904919
41   4.458811448348255    10.02560504518575      2.24849271186376
42   4.493122889418268    10.16522946629327      2.262397382950141
43   4.526474326127627    10.30194935868578      2.275932351857398
44   4.558910715791747    10.43586042057516      2.289112700633962
45   4.590474115000052    10.56705429924105      2.301952703471661
46   4.62120391564534    10.6956187836084      2.314465879204723
47   4.651137057938973    10.82163798838195      2.326665039876781
48   4.680308223012721    10.94519252981278      2.338562335701759
49   4.708750007375382    11.06635969324407      2.350169296715833
50   4.73649308120685    11.1852135926374      2.361496871391487
51   4.763566332226935    11.30182532231984      2.372555462460898
52   4.789996996664597    11.41626310121996      2.383354960174174
53   4.815810778670352    11.52859240987866      2.393904773198276
54   4.841031959356151    11.63887612053186      2.404213857344543
55   4.865683496509348    11.74717462056706      2.414290742296434
56   4.889787115907601    11.85354592965637      2.424143556494324
57   4.913363395057002    11.95804581086575      2.433780050320707
58   4.93643184008435    12.06072787603474      2.443207617716981
59   4.959010956434438    12.16164368571353      2.452433316351822
60   4.981118313952936    12.26084284393527      2.461463886451085
61   5.002770606873656    12.35837308809172      2.470305768389958
62   5.023983709174518    12.45428037416987      2.478965119139729
63   5.044772725718442    12.54860895759598      2.487447827653899
64   5.06515203955289    12.64140146992287      2.495759529271456
65   5.085135355704092    12.73269899158463      2.503905619208763
66   5.104735741768568    12.82254112093224      2.511891265205786
67   5.123965665574847    12.91096603975298      2.519721419387091
68   5.142837030161822    12.99801057546534      2.527400829393256
69   5.161361206296602    13.08371026017186      2.534934048833937
70   5.179549062733651    13.16809938674168      2.542325447109839
71   5.197410994398156    13.25121106208583      2.549579218647163
72   5.214956948659732    13.33307725777886      2.556699391584722
73   5.232196449847382    13.41372885817208      2.563689835950893
74   5.2491386221431    13.49319570613536      2.5705542713647
75   5.265792210979244    13.57150664655682      2.577296274292794
76   5.282165603053812    13.64868956772222      2.583919284891677
77   5.29826684506785    13.72477144068912      2.59042661346238
78   5.314103661280238    13.79977835676407      2.596821446542787
79   5.329683469967026    13.87373556318502      2.603106852660962
80   5.345013398865164    13.9466674971053      2.609285787771161
81   5.360100299673816    14.01859781796976      2.615361100392628
82   5.374950761680474    14.08954943836886      2.621335536469878
83   5.389571124573571    14.15954455345108      2.627211743971818
84   5.403967490498364    14.22860466896988      2.632992277245858
85   5.418145735408332    14.29675062803671      2.638679601142042
86   5.432111519760208    14.36400263664786      2.644276094921177
87   5.445870298597022    14.43038028804854      2.649784055960005
88   5.459427331060111    14.49590258599466      2.655205703265556
89   5.472787689367907    14.56058796696874      2.660543180810006
90   5.485956267296481    14.62445432140357      2.665798560696622
91   5.498937788194169    14.68751901396411      2.67097384616665
92   5.511736812560254    14.74979890293522      2.67607097445638
93   5.524357745215423    14.81131035876027      2.681091819512984
94   5.536804842089758    14.87206928177309      2.686038194577204
95   5.549082216652123    14.9320911191635      2.690911854640413
96   5.561193846003126    14.99139088121412      2.695714498783125
97   5.573143576652275    15.04998315684457      2.700447772401537
98   5.584935129998476    15.10788212849676      2.70511326932832
99   5.59657210753173    15.16510158639337      2.709712533853454
100   5.60805799577263    15.22165494219984      2.714247062650558
101   5.61939617096513    15.2775552421185      2.718718306613819
102   5.630589903537044    15.332815179442      2.723127672610321
103   5.641642362341722    15.38744710659176      2.72747652515229
104   5.652556618693491    15.44146304666565      2.731766187993482
105   5.663335650208609    15.49487470451806      2.735997945653691
106   5.673982344462718    15.54769347739415      2.740173044875133
107   5.684499502475065    15.59993046513886      2.744292696014233
108   5.694889842029118    15.6515964800005      2.74835807437212
109   5.705156000838572    15.70270205604725      2.752370321466966
110   5.715300539567187    15.75325745821449      2.756330546251111
111   5.725325944710379    15.80327269099944      2.760239826275753
112   5.735234631345977    15.85275750681911      2.764099208805813
113   5.745028945761108    15.9017214140467      2.767909711887452
114   5.754711167961762    15.95017368474058      2.771672325370572
115   5.764283514071177    15.99812336207967      2.775388011888501


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The Purdue solution manages to bound each term of the fancy sum by $C$ times the corresponding term in the harmonic sum (and this leads to bounding the fancy sum itself). I tried the other extreme as your investigation and looked numerically at $a_n=n!$, and the ratio of terms seems unbounded (although the sum itself is bounded with $C<1.8$). Either the ratio is unbounded or a huge $C$ is needed. If there is a single $C$ at all that works for all sequences, it must be enormous. – alex.jordan Sep 24 '12 at 21:42
Also, if $a_n$ is such that the harmonic series "slowly" converges (like $a_n=n^k$ or the $a_n$ that you tried), then the limit comparison test will show the fancy sum and the harmonic sum as having the same convergence status. – alex.jordan Sep 24 '12 at 21:51
Ammending my first comment: with $a_n=n!$ the ratio of terms seems to grow quadratically. The second differences are almost constant at 2. So I doubt there is a $C$ that bounds the sum term-wise, as in the Purdue solution. – alex.jordan Sep 24 '12 at 22:01
@alex.jordan, good to know. It is hard to see a single mechanism that accounts for these behaviours. But any splitting according to growth rates is going to do badly once oscillatory $a_n$ are allowed, or rather oscillatory $a_{n+1} - a_n.$ – Will Jagy Sep 24 '12 at 23:27
In an effort to see what the sharpest bound might be, I'm having fun setting $a_1=1$, and then recursively defining each $a_n$ to maximize the ratio you were computing. The behavior of this $\{a_n\}\approx\{1,3.43,9.53, 23.9,\cdots\}$ is interesting. – alex.jordan Sep 25 '12 at 3:26

As suggested by Alex Jordan, here is the same program with $$a_n = n!$$ We know that the sum is $e-1.$ I'm not sure what the fancy sum is, but it also converges rapidly. I stopped this at $n=69$ because my calculator says $70! > 10^{100}.$

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jagy@phobeusjunior:~$jagy@phobeusjunior:~$ ./series_potato
n    harmonic             fancy                  ratio
1   1                    1                      1
2   1.5                  1.888888888888889      1.259259259259259
3   1.666666666666667    2.555555555555555      1.533333333333333
4   1.708333333333333    2.908172635445363      1.702344957333871
5   1.716666666666667    3.036328472943761      1.768735032782773
6   1.718055555555556    3.070338463378451      1.78710080325989
7   1.718253968253968    3.077401815520791      1.791005213651823
8   1.71827876984127    3.07860906311077      1.791681953560518
9   1.718281525573192    3.078784678063843      1.791781283941121
10   1.718281801146385    3.078806934166605      1.791793949113888
11   1.718281826198493    3.07880943411009      1.79179537789887
12   1.718281828286169    3.078809686323769      1.791795522504363
13   1.718281828446759    3.078809709421133      1.79179553577903
14   1.71828182845823    3.078809711357875      1.791795536894207
15   1.718281828458995    3.078809711507647      1.791795536980574
16   1.718281828459042    3.078809711518395      1.791795536986779
17   1.718281828459045    3.078809711519114      1.791795536987195
18   1.718281828459046    3.078809711519159      1.791795536987221
19   1.718281828459046    3.078809711519162      1.791795536987222
20   1.718281828459046    3.078809711519162      1.791795536987222
21   1.718281828459046    3.078809711519162      1.791795536987222
22   1.718281828459046    3.078809711519162      1.791795536987222
23   1.718281828459046    3.078809711519162      1.791795536987222
24   1.718281828459046    3.078809711519162      1.791795536987222
25   1.718281828459046    3.078809711519162      1.791795536987222
26   1.718281828459046    3.078809711519162      1.791795536987222
27   1.718281828459046    3.078809711519162      1.791795536987222
28   1.718281828459046    3.078809711519162      1.791795536987222
29   1.718281828459046    3.078809711519162      1.791795536987222
30   1.718281828459046    3.078809711519162      1.791795536987222
31   1.718281828459046    3.078809711519162      1.791795536987222
32   1.718281828459046    3.078809711519162      1.791795536987222
33   1.718281828459046    3.078809711519162      1.791795536987222
34   1.718281828459046    3.078809711519162      1.791795536987222
35   1.718281828459046    3.078809711519162      1.791795536987222
36   1.718281828459046    3.078809711519162      1.791795536987222
37   1.718281828459046    3.078809711519162      1.791795536987222
38   1.718281828459046    3.078809711519162      1.791795536987222
39   1.718281828459046    3.078809711519162      1.791795536987222
40   1.718281828459046    3.078809711519162      1.791795536987222
41   1.718281828459046    3.078809711519162      1.791795536987222
42   1.718281828459046    3.078809711519162      1.791795536987222
43   1.718281828459046    3.078809711519162      1.791795536987222
44   1.718281828459046    3.078809711519162      1.791795536987222
45   1.718281828459046    3.078809711519162      1.791795536987222
46   1.718281828459046    3.078809711519162      1.791795536987222
47   1.718281828459046    3.078809711519162      1.791795536987222
48   1.718281828459046    3.078809711519162      1.791795536987222
49   1.718281828459046    3.078809711519162      1.791795536987222
50   1.718281828459046    3.078809711519162      1.791795536987222
51   1.718281828459046    3.078809711519162      1.791795536987222
52   1.718281828459046    3.078809711519162      1.791795536987222
53   1.718281828459046    3.078809711519162      1.791795536987222
54   1.718281828459046    3.078809711519162      1.791795536987222
55   1.718281828459046    3.078809711519162      1.791795536987222
56   1.718281828459046    3.078809711519162      1.791795536987222
57   1.718281828459046    3.078809711519162      1.791795536987222
58   1.718281828459046    3.078809711519162      1.791795536987222
59   1.718281828459046    3.078809711519162      1.791795536987222
60   1.718281828459046    3.078809711519162      1.791795536987222
61   1.718281828459046    3.078809711519162      1.791795536987222
62   1.718281828459046    3.078809711519162      1.791795536987222
63   1.718281828459046    3.078809711519162      1.791795536987222
64   1.718281828459046    3.078809711519162      1.791795536987222
65   1.718281828459046    3.078809711519162      1.791795536987222
66   1.718281828459046    3.078809711519162      1.791795536987222
67   1.718281828459046    3.078809711519162      1.791795536987222
68   1.718281828459046    3.078809711519162      1.791795536987222
69   1.718281828459046    3.078809711519162      1.791795536987222
jagy@phobeusjunior:~$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - I'm sorry that this is not an answer, but it's worthwhile information that might help. If we apply the Limit Comparison Test to the two series, putting the "harmonic" series below, we have the ratio $$\left(\frac{n\,a_n}{a_1+\cdots+a_n}\right)^2$$ Now if$a_n=f(n)$where$f$is an increasing continuous function, but not one that increases too quickly (as defined below when it matters) then $$\left(\frac{n\,a_n}{a_1+\cdots+a_n}\right)^2<\left(\frac{n\,f(n)}{\int_0^nf(x)\,dx}\right)^2$$ And so if$f$is slow-growing, as defined by$\int_0^nf(x)\,dx>C\,n\,f(n)$, then this ratio is bounded. So the Limit Comparison Test would give the convergence of$\sum\frac{n^2a_n}{(a_1+\cdots+a_n)^2}$. I've found this problem to be much harder to tackle for quickly growing$a_n$, which is funny, since for these the series$\sum\frac{1}{a_n}$has "more room" between it and a divergent series. If$a_n$is all-the-time "quickly growing", then this lends itself to a direct examination of$\sum\frac{n^2a_n}{(a_1+\cdots+a_n)^2}$, where the denominator can be shown to be larger enough than the numerator to give convergence. But I think the real problem with any continued approach like this will be sequences that go back-and-forth between slowly growing and quickly growing. And of course there is the concern that$a_n$might not even be increasing. - Alex, I get a different sense of$a_n = n!$from what you indicated. Posted as a separate answer. – Will Jagy Sep 25 '12 at 0:00 @Will Hi Will. For reasons no longer clear to me I was looking at the ratio of the terms as well as the ratio of the partial sums. My comments refer to the ratio of the terms. Does that clear up whatever I was saying? (I'm no longer sure why I was looking at that ratio though!) – alex.jordan Sep 25 '12 at 2:38 Alex, I thought it was something like that. I went through the accepted answer line by line, more detail could be typed in at various points but it's correct. – Will Jagy Sep 25 '12 at 2:49 Assume this fact$(\clubsuit)$: Prove that$\sum_{k=1}^n \frac{2k+1}{a_1+a_2+...+a_k}<4\sum_{k=1}^n\frac1{a_k}.$. If you define$P_N=\sum_{n=1}^N a_n\,,\; C=\sum_{n=1}^{+\infty}\frac{1}{a_n}$and$S_N=\sum_{n=1}^{N}\frac{n^2 a_n}{P_n^2}$you have: $$S_N < \frac{1}{a_1} + \sum_{n=2}^{N}\frac{n^2(P_n-P_{n-1})}{P_n P_{n-1}} = \frac{5}{a_1}+\sum_{n=2}^{N-1}\frac{2n+1}{P_n}-\frac{N^2}{P_N},$$ so, in virtue of$(\clubsuit)\$, you have:

$$S_N < \frac{2}{a_1} + 4C.$$

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