Cauchy product summation converges

calculuscauchy-productcauchy-sequencesconvergence-divergencesummation

I had a previous question here, which I'm quoting:

How can I prove that the following summation converges?
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1) (n-k+1)}$$

I tried to prove that by proving that the following summation in in
absolute value converges so the original one converges too, but that's
incorrect.

And the answer was:

$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1) (n-k+1)} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{(k+1)}\cdot\frac{(-1)^{n-k}}{(n-k+1)}$$

Then this is the Cauchy product (discrete convolution) of
$\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)}$ with itself. We know that
$\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k}$, by changing the
index $k \rightarrow k+1$ we get $\ln(1+x)=\sum_{k=0}^\infty \frac{(-1)^{k}x^{k+1}}{k+1}$ and we have
$\ln(2)=\ln(1+1)=\sum_{k=0}^\infty \frac{(-1)^{k}}{k+1}$ and we can
conclude that

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

While this answer finds the value to which the summations converges (if it converges) it doesn't prove that the summation converges at all.
According to what I read on Cauchy product (From wikipedia):

Let $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$ be real or complex sequences. It was proved by
Franz Mertens[citation needed] that, if the series $a_n$ converges to A
and the series $b_n$ converges to B and at least one of them converges
absolutely, then their Cauchy product converges to AB.

But after doing some checks I can confirm that non of them converges absolutely

Best Answer

I'll start much as in Exodd's answer, but diverge a bit.

The series is $\sum_{n=0}^\infty(-1)^na_n,$ where $$ a_n = \sum_{k=0}^n\frac1{(k + 1)(n - k + 1)} = \frac1{n + 2}\sum_{k=0}^n\left(\frac1{k + 1} + \frac1{n - k + 1}\right) = \frac{2H_{n + 1}}{n + 2}, $$ where $$ H_m = \sum_{k=1}^m\frac1k \quad (m \geqslant 1). $$ By the alternating series test, $\sum_{n=0}^\infty(-1)^na_n$ converges if the sequence $(a_n)$ decreases to zero (it doesn't have to be strictly decreasing) as $n$ tends to infinity. Therefore, it is enough to prove that $$ \frac{H_m}{m + 1} \searrow 0 \text{ as } m \to \infty \quad (m \geqslant 1). $$ It is straightforward to prove that the sequence $\left(\frac{H_m}{m + 1}\right)$ is decreasing: \begin{gather*} mH_m = mH_{m - 1} + 1 \leqslant (m + 1)H_{m - 1} \quad (m \geqslant 2), \\ \therefore\ \frac{H_m}{m + 1} \leqslant \frac{H_{m - 1}}m. \end{gather*} Here is a simple proof from first principles that $\frac{H_m}{m + 1} \to 0$ as $m \to \infty$: \begin{gather*} H_m \leqslant \sum_{k=1}^m\frac1{\sqrt{k}} < 2\sum_{k=1}^m\frac1{\sqrt{k} + \sqrt{k - 1}} = 2\sum_{k=1}^m\left(\sqrt{k} - \sqrt{k - 1}\right) = 2\sqrt{m}, \\ \therefore\ \frac{H_m}{m + 1} < \frac2{\sqrt{m + 1}} \to 0 \text{ as } m \to \infty. \end{gather*}

Optional Extra

It seems a shame not to evaluate the sum, and I shall prove that it is $(\log2)^2,$ as expected.

For each finite set $F$ of ordered pairs of positive integers, define the finite sum: $$ \mu(F) = \sum_{(i, j) \in F}\frac{(-1)^{i + j}}{ij}. $$ For each positive integer $p,$ the $p^\text{th}$ partial sum of the given series is: \begin{gather*} \sum_{n=0}^{p-1}\sum_{k=0}^n\frac{(-1)^n}{(k + 1)(n - k + 1)} = \mu(F_p), \text{ where:} \\ F_p = \{ (i, j) \colon i + j \leqslant p + 1 \}). \end{gather*} We have proved that $\mu(F_p)$ tends to a limit $l$ as $p \to \infty.$ Therefore: $$ \mu(F_{4r - 1}) \to l \text{ as } r \to \infty. $$

For each positive integer $r,$ define this "square" set of pairs of integers: $$ K_r = \{ (i, j) \colon i \leqslant 2r \text{ and } j \leqslant 2r \} \subset F_{4r - 1}. $$ Then: $$ \mu(K_r) = \left(\sum_{i=1}^{2r}\frac{(-1)^{i-1}}{i}\right)\left(\sum_{j=1}^{2r}\frac{(-1)^{j-1}}{j}\right) \to (\log2)^2 \text{ as } r \to \infty. $$ We have $F_{4r-1} = K_r \sqcup L_r \sqcup M_r,$ a disjoint union, where: \begin{align*} L_r & = \{ (i, j) \colon i < 2r < j \text{ and } i + j \leqslant 4r \}, \\ M_r & = \{ (i, j) \colon j < 2r < i \text{ and } i + j \leqslant 4r \}. \end{align*} Clearly $\mu(M_r) = \mu(L_r),$ so we have: $$ \mu(F_{4r-1}) = \mu(K_r) + 2\mu(L_r). $$ We now only need to show that $\mu(L_r) \to 0$ as $r \to \infty,$ and it will follow that $l = (\log2)^2.$ \begin{align*} \mu(L_r) & = \sum_{j=2r+1}^{4r-1}\frac{(-1)^{j-1}}{j}\sum_{i=1}^{4r-j}\frac{(-1)^{i-1}}{i} \\ & = \sum_{j=2r+1}^{4r-1}\frac{(-1)^{j-1}}{j}\sum_{i=1}^{4r-j+1}\frac{(-1)^{i-1}}{i} + \sum_{j=2r+1}^{4r-1}\frac1{j(4r-j+1)}. \end{align*} When $j$ is odd, $4r-j+1$ is even. When $j$ is even, $4r-j+1$ is odd. By the theory of convergent alternating series with non-zero terms, if the first term of the series is positive, the odd-numbered partial sums are greater than the infinite sum, and the even-numbered partial sums are less than the infinite sum. Hence: \begin{align*} \mu(L_r) & < (\log2)\!\sum_{j=2r+1}^{4r-1}\frac{(-1)^{j-1}}{j} + \frac1{4r+1}\sum_{j=2r+1}^{4r-1}\left(\frac1j + \frac1{4r-j+1}\right) \\ & = (\log2)\left(\sum_{j=1}^{4r-1}\frac{(-1)^{j-1}}{j} - \sum_{j=1}^{2r}\frac{(-1)^{j-1}}{j}\right) + \frac{H_{4r-1} - 1}{4r+1} \\ & \to 0 \text{ as } r \to \infty. \end{align*} The first term tends to zero because of the convergence of the series $\sum_{j=1}^\infty\frac{(-1)^{j-1}}{j},$ and the second term tends to zero because, as shown above, $H_{4r-1} < 2\sqrt{4r+1}.$ $\ \square$

Corollary. $$ \sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{n+1} = \frac{(\log2)^2}{2}. $$ (I've seen many similar but more complicated equations posted on Maths.SE, so this one is bound to have been posted before, probably with a much nicer proof, but it seemed worth mentioning.)

[Update]

This must be the simplest equation that comes under the heading of Euler sums. In spite of much searching, however, I haven't found it written down anywhere in such a simple form.

It is a special case of equation (2.33) in Ce Xu, Explicit evaluation of harmonic sums (2017).

Euler's equation $\sum_{n=1}^\infty\frac1{n^2} = \frac{\pi^2}6$ gives $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12},$ whence: $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}n = \frac{\pi^2}{12} - \frac{(\log2)^2}{2}. $$ Simple proofs of the latter more complex equation (it's still very simple, by the standards of results in this area!) have been requested more than once in Maths.SE:

I'm reluctant to make a fool of myself by straying ignorantly into this area of exquisite expertise. (Browse the tag just given, and you'll see what I mean.) If nevertheless it is worth polishing and simplifying my proof, I'll do so in a separate question.

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