46
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Let $L$ be your limit. Numerically, I get $(1-\sqrt{5})/2 < -0.61803 < -0.617 < L$. $\endgroup$
    – Gerald Edgar
    Commented Jun 8, 2016 at 14:22
  • 11
    $\begingroup$ It seems to be $-\pi^2/16$ which is pretty close to $(1-\sqrt{5})/2$. $\endgroup$
    – Lucia
    Commented Jun 8, 2016 at 14:22
  • 7
    $\begingroup$ 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). $\endgroup$
    – Lucia
    Commented Jun 8, 2016 at 15:21
  • 2
    $\begingroup$ @GHfromMO yes, it is a review queue where more experienced users (with >3k reputation) decide which questions to close or leave open. $\endgroup$
    – Glorfindel
    Commented Jun 8, 2016 at 19:05
  • 9
    $\begingroup$ 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... $\endgroup$
    – Gottfried Helms
    Commented Jun 9, 2016 at 6:44

2 Answers 2

Reset to default
97
$\begingroup$

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.)

Improve this answer
$\endgroup$
5
  • 48
    $\begingroup$ We may do the change of variables $y=z^2$, $xz=t$, we get $2\int_{t<z} dtdz/(1+t^2)(1+z^2)=\int_0^1\int_0^1 dtdz/(1+t^2)(1+z^2)=(\pi/4)^2$ $\endgroup$
    – Fedor Petrov
    Commented Jun 8, 2016 at 14:43
  • 5
    $\begingroup$ @FedorPetrov: Nicely done! $\endgroup$
    – Lucia
    Commented Jun 8, 2016 at 14:44
  • 1
    $\begingroup$ Gee you must be good at those Facebook math questions :D $\endgroup$
    – Xela
    Commented Jun 13, 2016 at 0:12
  • $\begingroup$ 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? $\endgroup$
    – John Bentin
    Commented Jul 17, 2016 at 14:09
  • $\begingroup$ @JohnBentin: Yes, that's right. Now fixed, thanks! $\endgroup$
    – Lucia
    Commented Jul 17, 2016 at 14:46
16
$\begingroup$

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 $$\sum_{n\geq1}\frac{(-1)^n}{n}\sum_{k=1}^n\frac1{2k-1}=-\frac{\pi^2}{16}.$$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .