# Is the following identity true?

Calculation suggests the following identity: $$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{(-1)^k}{k}\sum_{j=1}^k\frac{1}{2j-1}=\frac{1-\sqrt{5}}{2}.$$

I have verified this identity for $n$ up to $5000$ via Maple and find that the left-hand side approaches $\frac{1-\sqrt{5}}{2}$. However, this double summation has slow rate of convergence and I am unsure it is true.

So I want to ask if it is true. If so, how to prove it?

• Let $L$ be your limit. Numerically, I get $(1-\sqrt{5})/2 < -0.61803 < -0.617 < L$. Jun 8, 2016 at 14:22
• It seems to be $-\pi^2/16$ which is pretty close to $(1-\sqrt{5})/2$. Jun 8, 2016 at 14:22
• Just since this came up on my Close queue: while the problem may not look research level, very similar double integrals occur in Beukers's proof of irrationality of $\zeta(2)$ and $\zeta(3)$, which is certainly research level" and is what piqued my interest in this (plus the fact that the answer is another pretty constant very close to the one guessed by OP). Jun 8, 2016 at 15:21
• @GHfromMO yes, it is a review queue where more experienced users (with >3k reputation) decide which questions to close or leave open. Jun 8, 2016 at 19:05
• Just a comment on the numerical approach: Euler-summation can often improve convergence dramatically. For your problem I arrived with the first 36 partial sums 20 correct digits. Here is the protocol of index and partial sums:$$\small \begin{array} {}...& ...\\33& -0.61685027506808491388 \\ 34 &-0.61685027506808491375 \\ 35 &-0.61685027506808491370 \\ 36 &-0.61685027506808491369 \\ 37 &-0.61685027506808491368 \\ 38 &-0.61685027506808491368 \\ ...& ...\\ \end{array}$$ (I used Euler-summation to order 0.58) With W/A you'd found the correct value easily... Jun 9, 2016 at 6:44

You can evaluate this by using generating functions and integrating. The answer is $-\pi^2/16 = -0.61685 \ldots$ which is pretty close to $(1-\sqrt{5})/2=-0.61803\ldots$.

Here's a sketch: the sum is $$\sum_{k=1}^{\infty} \frac{(-1)^k}{k} \int_0^1 (1+x^2+ \ldots +x^{2k-2}) dx = \int_0^1 \sum_{j=0}^{\infty} x^{2j} \sum_{k=j+1}^{\infty} \frac{(-1)^k}{k} dx$$ which is $$= \int_0^1 \sum_{j=0}^{\infty} x^{2j} \Big(\int_0^1 \sum_{k=j}^{\infty} -(-y)^{k} dy \Big) dx = - \int_0^1 \int_0^1 \sum_{j=0}^{\infty} \frac{(x^2 y)^j}{1+y} dy dx,$$ which is $$= - \int_0^1\int_0^1 \frac{dx dy}{(1+x^2y)(1+y)}.$$ The integral in $y$ can be done easily: $$\int_0^1 \frac{1}{1-x^2} \Big( \frac{1}{1+y}- \frac{x^2}{1+x^2y}\Big)dy = \log \Big(\frac{2}{1+x^2}\Big) \frac{1}{1-x^2}.$$ We're left with $$- \int_0^1 \log \frac{2}{1+x^2} \frac{dx}{1-x^2},$$ which WolframAlpha evaluates as $-\pi^2/16$. (This doesn't look too bad to do by hand, but I don't see a reason to do one variable integrals that a computer can recognize at once.)

• We may do the change of variables $y=z^2$, $xz=t$, we get $2\int_{t<z} dtdz/(1+t^2)(1+z^2)=\int_0^1\int_0^1 dtdz/(1+t^2)(1+z^2)=(\pi/4)^2$ Jun 8, 2016 at 14:43
• @FedorPetrov: Nicely done! Jun 8, 2016 at 14:44
• Gee you must be good at those Facebook math questions :D
– Xela
Jun 13, 2016 at 0:12
• In the second display line, should the terminal $dy$ on the LHS be $dx$, and is there a missing $dx$ at the end of the line? Jul 17, 2016 at 14:09
• @JohnBentin: Yes, that's right. Now fixed, thanks! Jul 17, 2016 at 14:46

Start with $$\sum_{n\geq1}\frac{z^{2n-1}}{2n-1}=\frac12\log\left(\frac{1+z}{1-z}\right).$$ Since $$\frac1{1-z^2}\sum_{n\geq1}\frac{z^{2n-1}}{2n-1}=\sum_{n\geq1}z^{2n-1}\sum_{k=1}^n\frac1{2k-1}$$, we have $$\int_0^z\left(\sum_{n\geq1}z^{2n-1}\sum_{k=1}^n\frac1{2k-1}\right)dz=\int_0^z \frac1{2(1-z^2)}\log\left(\frac{1+z}{1-z}\right)\,dz.$$ Therefore, $$\sum_{n\geq1}\frac{z^{2n}}{2n}\sum_{k=1}^n\frac1{2k-1}= \frac18\log^2\left(\frac{1+z}{1-z}\right).$$ Choose $$z=i=\sqrt{-1}$$ so that $$\sum_{n\geq1}\frac{(-1)^n}{2n}\sum_{k=1}^n\frac1{2k-1}= \frac18\log^2\left(\frac{1+i}{1-i}\right)=-\frac{\pi^2}{32}.$$ It follows that $$\sum_{n\geq1}\frac{(-1)^n}{n}\sum_{k=1}^n\frac1{2k-1}=-\frac{\pi^2}{16}.$$