# Is there a closed form for the alternating series of inverse harmonic numbers?

Let $H_n=\sum _{k=1}^n \frac{1}{k}$ be the n-th harmonic number. Since $H_{k+1}>H_k$ for $k=1, 2, 3, ...$ the sequence $\frac{1}{H_k}$ is monotonic decreasing as $n \to \infty$, and the Leibniz criterion tells us that the alternating series

$$s_{H}=\sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{H_k}\tag{1}$$

converges.

Its numerical value is

$$N(s_{H}) = 0.626332...$$

The natural question arises:

Question 1

Is there a closed form for $s_{H}$, i.e. an expression in terms of known constants (or is it even a new constant)? A possible selection of these constants might be those arising in the power series expansion of the zeta function, i.e.

$$const=\left\{\gamma ,\log (\pi ),\log (2),\gamma _1,\gamma _2,\zeta (3), ...\right\}$$

Question 2

A variation of the question replaces the harmonic number by the logarithm, starting at $k=2$, and asks for a closed form of the series

$$s_{L}=\sum _{k=2}^{\infty } \frac{(-1)^k}{\log (k)}\tag{2}$$

here

$$N(s_{L})=0.924299 ...$$

Solution attempts

For question 1 I have no idea to find a closed form, but the numerical value can be found to high precision.

Mathematica gives the first 100 Digits with this command:

NSum[(-1)^(n + 1)/HarmonicNumber[n], {n, 1, Infinity}, WorkingPrecision -> 100, Method -> "AlternatingSigns"]

$N(s_{H})=0.626332482737912354708657266227986063950088333562581965723069813694423327263764315345087698850095778034583404083688989609231701677593263$

I cannot tell if all these digits are correct.

My attempt to solve question 2 starts with the replacement

$$\frac{1}{\log (k)}=\int_0^{\infty } \exp (-t \log (k)) \, dt=\int_0^{\infty } k^{-t} \, dt$$

The summation under the integral is just the definition of the alternating zeta function starting at k = 2 which can be written as

$$\sum _{k=1}^{\infty } (-1)^k k^{-t}= (1-\zeta (t))+2^{1-t} \zeta (t)$$

Where $\zeta (t)=\sum _{k=1}^{\infty } k^{-t}$ is the Riemann zeta function.

Hence we find

$$s_{L}=\int_0^{\infty } ( (1-\zeta (t))+2^{1-t} \zeta (t) )\, dt\tag{3}$$

The integrand is well behaved in the region of integration (it resembles the decaying exponential).

But still I am stuck here (and Mathematica as well).

Table lookups

Stimulated by a comment of "J. M. is not a mathematician" I looked up the constants $s_{H}$ and $s_{L}$ defined here in the available tables.

The results are:

The Inverse Symbolic Calculator [1] could not identify the two constants.

The Online-Encyclopedia of integer sequences, searched for the sequence of the decimal digits, does not find $s_{H}$ but it does find $s_{L}$, and what's more, it contains the series of question 2: A099769 Decimal expansion of Sum_{n >= 2} (-1)^n/log(n). [2]

Here I.V.Blagouchine gives the following interesting integral representation without proof

$$s_{LB}=\int_0^{\infty } \frac{8 \tan ^{-1}(x)}{\sinh (2 \pi x) \left(\log ^2\left(4 x^2+4\right)+4 \tan ^{-1}(x)^2\right)} \, dx+\frac{1}{2 \log (2)}\tag{4}$$

Mathematica gives the first 100 digits with this command:

NSum[(-1)^n/Log[n], {n, 2, Infinity}, WorkingPrecision -> 100, Method -> "AlternatingSigns"]

$N(s_{L})=0.9242998972229388559595701813595900537733193978869190747796304372507005417114 3468979899134744193228$

Ths coincides with the digits given in the comment of robjohn and those of Ref. [2].

References

• FWIW, the ISC does not seem to know any of your constants either. – J. M. is a poor mathematician Sep 14 '17 at 9:55
• The sum with log can be computed to great precision with the Euler-Maclaurin Sum Formula: $$0.9242998972229388559595701813595900537733$$ However, the sum with the Harmonic Numbers is not as compliant (so far). – robjohn Sep 15 '17 at 15:38
• @robjohn Could you please explain this calculation in some detail, maybe as a solution? – Dr. Wolfgang Hintze Sep 15 '17 at 22:01

This is not an answer, but an attempt to accelerate the series $(1)$ to get a good numerical value.

First, let's rewrite the series as absolutely convergent, using the definition of harmonic numbers:

$$\frac{1}{H_{2n-1}}-\frac{1}{H_{2n}}=\frac{1}{H_{2n-1}}-\frac{1}{H_{2n-1}+1/2n}=\frac{1}{2n H^2_{2n-1} \left(1+\frac{1}{2n H_{2n-1}} \right)}$$

For $n \geq 1$ the second term in the brackets is less than $1$, which makes it possible to use the geometric series expansion:

$$\frac{1}{2n H^2_{2n-1} \left(1+\frac{1}{2n H_{2n-1}} \right)}=\frac{1}{2n H^2_{2n-1}} \sum_{k=0}^\infty \frac{(-1)^k}{(2n)^k H_{2n-1}^k}$$

Shifting the index $k \to k+2$ we can rewrite the original series as:

$$s_H=\sum_{n=1}^\infty \left(\frac{1}{H_{2n-1}}-\frac{1}{H_{2n}} \right)=\sum_{k=2}^\infty \frac{(-1)^k}{2^{k-1}} \sum_{n=1}^\infty \frac{1}{n^{k-1} H_{2n-1}^k} \tag{a}$$

This is again an alternating series (the outer one), and obviously we do not know the closed form of the inner series. However, we can make some conclusions.

Firstly:

$$\lim_{k \to \infty} \sum_{n=1}^\infty \frac{1}{n^{k-1} H_{2n-1}^k}=1$$

Which means that for some large $K$ we can use the following approximation:

$$s_H \approx \sum_{k=2}^K \frac{(-1)^k}{2^{k-1}} \sum_{n=1}^\infty \frac{1}{n^{k-1} H_{2n-1}^k}+\sum_{k=K+1}^\infty \frac{(-1)^k}{2^{k-1}} \tag{b}$$

The last part is just a geometric series, which is computed exactly and has a rational value for any $K$.

Now we need to evaluate the inner series for the first few values of $k$.

For some large $N$ we can use the asymptotic expansion for harmonic numbers for $n > N$:

$$H_{2n-1} \approx \gamma+ \log (2n-1)+\frac{1}{2(2n-1)}-\frac{1}{12(2n-1)^2}+\dots$$

Using the terms up to $\frac{1}{2(2n-1)}$ and $\frac{1}{12(2n-1)^2}$ and comparing the results we can estimate the error quite easily.

Then we can write:

$$\sum_{n=1}^\infty \frac{1}{n^{k-1} H_{2n-1}^k}=\sum_{n=1}^N \frac{1}{n^{k-1} H_{2n-1}^k}+\sum_{n=N+1}^\infty \frac{1}{n^{k-1} \left(\gamma+ \log (2n-1)+\frac{1}{2(2n-1)} \right)^k} \tag{c}$$

Again, the first part can be computed exactly using the known values of the harmonic numbers and the second part can be found numerically with high precision, using the usual methods.

For $k=1$ and choosing $N=5$ we already obtain $5$ correct digits using the above approximation.

And using $K=5$ in $(b)$ after computing the inner series for $k=2,3,4,5$ we also have at least $5$ correct digits for the final expression.

While these results are not that impressive, I believe it's possible to get better numerical results by some more advanced method.

Update:

There exist very sharp inequalities for harmonic numbers, see this paper: https://www.sciencedirect.com/science/article/pii/S0723086906000168.

$$a-\ln \left(e^{1/(n+1)}-1 \right) \leq H_n < b-\ln \left(e^{1/(n+1)}-1 \right)$$

Where:

$$a=1+\ln \left(\sqrt{e}-1 \right)$$

$$b=\gamma$$

Using these inequalities, we can get very good estimates for the series $\sum_{n=1}^\infty \frac{1}{n^{k-1} H_{2n-1}^k}$.

I don't compute a closed form, but at the request of Dr. Wolfgang Hintze, I am supplying the computations involved in computing $47$ decimal places of $$\sum_{k=2}^\infty\frac{(-1)^k}{\log(k)}$$ The Euler-Maclaurin Sum Formula says that \newcommand{\li}{\operatorname{li}} \begin{align} \sum_{k=2}^n\frac1{\log(k)} &=C_1+\li(n)+\frac1{2\log(n)}-\frac1{12n\log(n)^2}\\ &+\frac1{n^3}\left(\frac1{360\log(n)^2}+\frac1{120\log(n)^3}+\frac1{120\log(n)^4}\right)\\ &-\frac1{n^5}\left(\scriptsize\frac1{1260\log(n)^2}+\frac5{1512\log(n)^3}+\frac1{144\log(n)^4}+\frac1{126\log(n)^5}+\frac1{252\log(n)^6}\right)\\ &+\frac1{n^7}\left(\tiny\frac1{1680\log(n)^2}+\frac7{2400\log(n)^3}+\frac{29}{3600\log(n)^4}+\frac7{480\log(n)^5}+\frac5{288\log(n)^6}+\frac1{80\log(n)^7}+\frac1{240\log(n)^8}\right)\\ &-\frac1{n^9}\left(\tiny\frac1{1188\log(n)^2}+\frac{761}{166320\log(n)^3}+\frac{29531}{1995840\log(n)^4}+\frac{89}{2640\log(n)^5}+\frac{1069}{19008\log(n)^6}+\frac3{44\log(n)^7}\right.\\ &\phantom{-\frac1{n^9}\left(\vphantom{\frac1{\log(n)}}\right.}\left.\tiny+\frac{91}{1584\log(n)^8}+\frac1{33\log(n)^9}+\frac1{132\log(n)^{10}}\right)\\ &+O\left(\frac1{n^{11}\log(n)^2}\right) \end{align} and \begin{align} \sum_{k=1}^n\frac1{\log(2k)} &=C_2+\frac12\li(2n)+\frac1{2\log(2n)}-\frac1{12n\log(2n)^2}\\ &+\frac1{n^3}\left(\frac1{360\log(2n)^2}+\frac1{120\log(2n)^3}+\frac1{120\log(2n)^4}\right)\\ &-\frac1{n^5}\left(\scriptsize\frac1{1260\log(2n)^2}+\frac5{1512\log(2n)^3}+\frac1{144\log(2n)^4}+\frac1{126\log(2n)^5}+\frac1{252\log(2n)^6}\right)\\ &+\frac1{n^7}\left(\tiny\frac1{1680\log(2n)^2}+\frac7{2400\log(2n)^3}+\frac{29}{3600\log(2n)^4}+\frac7{480\log(2n)^5}+\frac5{288\log(2n)^6}+\frac1{80\log(2n)^7}+\frac1{240\log(2n)^8}\right)\\ &-\frac1{n^9}\left(\tiny\frac1{1188\log(2n)^2}+\frac{761}{166320\log(2n)^3}+\frac{29531}{1995840\log(2n)^4}+\frac{89}{2640\log(2n)^5}+\frac{1069}{19008\log(2n)^6}+\frac3{44\log(2n)^7}\right.\\ &\phantom{-\frac1{n^9}\left(\vphantom{\frac1{\log(n)}}\right.}\left.\tiny+\frac{91}{1584\log(2n)^8}+\frac1{33\log(2n)^9}+\frac1{132\log(2n)^{10}}\right)\\ &+O\left(\frac1{n^{11}\log(n)^2}\right) \end{align} where the log-integral is defined as $$\li(x)=\int_0^x\frac{\mathrm{d}t}{\log(t)}$$ In any case, the log-integral disappears in the alternating sum \begin{align} \sum_{k=2}^{2n}\frac{(-1)^k}{\log(k)} &=2\sum_{k=1}^n\frac1{\log(2k)}-\sum_{k=2}^{2n}\frac1{\log(k)}\\ &=C_3+\frac1{2\log(2n)}-\frac1{8\,n\log(2n)^2}\\ &+\frac{15}{8\,n^3}\left(\frac1{360\log(2n)^2}+\frac1{120\log(2n)^3}+\frac1{120\log(2n)^4}\right)\\ &-\frac{63}{32\,n^5}\left(\scriptsize\frac1{1260\log(2n)^2}+\frac5{1512\log(2n)^3}+\frac1{144\log(2n)^4}+\frac1{126\log(2n)^5}+\frac1{252\log(2n)^6}\right)\\ &+\frac{255}{128\,n^7}\left(\tiny\frac1{1680\log(2n)^2}+\frac7{2400\log(2n)^3}+\frac{29}{3600\log(2n)^4}+\frac7{480\log(2n)^5}+\frac5{288\log(2n)^6}+\frac1{80\log(2n)^7}+\frac1{240\log(2n)^8}\right)\\ &-\frac{1023}{512\,n^9}\left(\tiny\frac1{1188\log(2n)^2}+\frac{761}{166320\log(2n)^3}+\frac{29531}{1995840\log(2n)^4}+\frac{89}{2640\log(2n)^5}+\frac{1069}{19008\log(2n)^6}+\frac3{44\log(2n)^7}\right.\\ &\phantom{-\frac1{512n^9}\left(\vphantom{\frac1{\log(n)}}\right.}\left.\tiny+\frac{91}{1584\log(2n)^8}+\frac1{33\log(2n)^9}+\frac1{132\log(2n)^{10}}\right)\\ &+O\left(\frac1{n^{11}\log(n)^2}\right) \end{align} Plug in $n=10000$ and we get $C_3$ to over $45$ places $$\sum_{k=2}^\infty\frac{(-1)^k}{\log(k)} =0.92429989722293885595957018135959005377331939789$$

• Thank you. Just one (maybe stupid) question: I don't understand your conclucding sentence. How is $C_{3}$ defined? – Dr. Wolfgang Hintze Sep 17 '17 at 8:38
• $C_3=2C_2-C_1$ is defined in the previous equation: $$\sum_{k=2}^{2n}\frac{(-1)^k}{\log(k)}=C_3+\frac1{2\log(2n)}-\frac1{8\,n\log(2n)^2}+\cdots$$ – robjohn Sep 17 '17 at 10:51
• @ robjohn Thank you. But I still have some basic methodological questions: I can reproduce the r.h.s of, say, your first sum when I put all constant terms appearing in the Euler-Maclaurin expansion into $C_1$. But now: how do I calculate $C_1$? And why would I want to calculate $C_1$ when I am actually interested in a good numerical approximation of $\sum_{k=2}^n \frac{1}{\log(k)}$ ? – Dr. Wolfgang Hintze Oct 2 '17 at 10:12
• To compute $C_1$, pick an $n$. Then compute both sides of the equation for $\sum\limits_{k=2}^n\frac1{\log(n)}$. This equation holds with an error of the order $\frac1{n^{11}\log(n)^2}$. If $n=10000$ is used, we get $\gt45$ places of accuracy. Then, once we have $C_1$, it can be used it to compute $\sum\limits_{k=2}^n\frac1{\log(n)}$ using the same equation for different $n$. This is the same reason we want to know $\gamma$ to compute $\sum\limits_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}+O\!\left(\frac1{n^2}\right)$. – robjohn Oct 2 '17 at 13:17
• @ robjohn Thank you again. I have now understood the procedure, and have calculated some asymptotics. But if we would be only interested in the values of convergent sums, these can be determined to high accuracy in Mathematica. The identication of these and of other constants appearing is of course a different question. – Dr. Wolfgang Hintze Oct 2 '17 at 20:48