Cauchy product of $\sum\limits_n^{\infty}\frac{1}{n}$ with itself Is the cauchy product of  $\sum\limits_n^{\infty}\frac{1}{n}$ with itself simply  $\sum\limits_n^{\infty}\frac{1}{n^2}$?
I can't apply the definition here https://proofwiki.org/wiki/Definition:Cauchy_Product because then $\infty$ appears :(
Please help.
 A: If you are talking about the Cauchy product formally, $$
\sum a_n\sum b_n=\sum c_n
$$
where 
$$
c_n=\sum_{k+l=n}a_kb_l
$$
Then $a_k=b_k=\dfrac{1}{k}$ and you just have
$$
c_n=\sum_{k+l=n}\frac{1}{k}\cdot\frac{1}{l}.
$$
A: Firstly, your series diverges. So it doesn't make sense to talk about the Cauchy product at all.
However, we can modify your question so that it does make sense, and consider the Cauchy product of 
$$ \sum_{n = 1}^\infty \frac{1}{n^s}$$
with itself, for real $s > 1$. Then your question might be phrased, is the Cauchy product of this with itself equal to
$$ \sum_{n = 1}^\infty \frac{1}{n^{2s}}?$$
And the answer is no. To see this, we can just write out the first two terms,
$$ \bigg(1 + \frac{1}{2^s}\bigg)\bigg(1 + \frac{1}{2^s}\bigg) = 1 + \frac{2}{2^s} + \frac{1}{4^s}.$$
This is not what we would get from squaring the denominator.
More generally, we have that
$$ \left( \sum_{n = 1}^\infty \frac{1}{n^s} \right)^2 = \sum_{n = 1}^\infty \frac{d(n)}{n^s},$$
where $d(n)$ denotes the number of positive divisors of $n$.
A: Notice that if $a_n, b_n \geq 0$ for all $n \geq 0$ and $c_n = \sum_{k=0}^{n} a_k b_{n-k}$ is the Cauchy product of $(a_n)$ and $(b_n)$, then we always have
$$ \sum_{n=1}^{\infty} c_n = \bigg( \sum_{n=0}^{\infty} a_n \bigg)\bigg( \sum_{n=0}^{\infty} b_n \bigg), $$
where we adopt the convention $\infty \cdot 0 = 0$ so that this identity is true in any cases. The proof follows by taking limit to the inequality
$$ \sum_{k=0}^{n} c_k
= \sum_{i+j\leq n} a_i b_j
\leq \bigg( \sum_{i=0}^{n} a_i \bigg)\bigg( \sum_{j=0}^{n} b_j \bigg)
\leq \sum_{i+j\leq 2n} a_i b_j
= \sum_{k=0}^{2n} c_k. $$
This immediately tells you that the sum of Cauchy product of $(\frac{1}{n} : n \geq 1)$ with itself diverges.
If you are just interested in the Cauchy product itself, here is a simple way of computing it:
$$ \sum_{k=1}^{n-1} \frac{1}{k(n-k)} = \sum_{k=1}^{n-1} \frac{1}{n}\left( \frac{1}{k} + \frac{1}{n-k}\right) = \frac{2}{n} \sum_{k=1}^{n-1} \frac{1}{k} = \frac{2H_{n-1}}{n}. $$
This is asymptotically $\frac{2}{n}\log n$, whose sum diverges.
A: That sum does not converge,
so let's look at
$f(x)
=\sum_{n=0}^{\infty} \dfrac{x^n}{n+1}
$.
We have
$f(x) = -\ln(1-x)/x
$,
so the Cauchy product would give
$f^2(x)
=\ln^2(1-x)/x^2
$.
This doesn't seems to help much,
so let's do the product.
$\begin{array}\\
f^2(x)
&=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty} \dfrac{x^i}{i+1}\dfrac{x^j}{j+1}\\
&=\sum_{n=0}^{\infty}x^n\sum_{j=0}^{n} \dfrac{1}{(j+1)(n-j+1)}\\
\end{array}
$
So, this depends on
$g(n)
= \sum_{j=0}^{n} \dfrac{1}{(j+1)(n-j+1)}
$.
Numerical experimentation
suggests that
$g(n)$
is decreasing and
is about
$\dfrac1{\sqrt{n}-1}$.
If this is true,
then
$f^2(x)
=\sum_{n=0}^{\infty} g(n) x^n
$
converges for
$|x| < 1$.
That's all I have for now.
