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I am trying to evaluate the series

$$S=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{k+n}^2}{k+n}(-1)^{k+n}$$

Where

$$H_{k}^2=(H_k)^2=\left(\sum_{n=1}^k \frac{1}{n}\right)^2$$

Attempt

I am first concerned about convergence

$$S=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{H_{k+n}^2}{k+n}(-1)^{k+n}-\sum_{k=1}^\infty\frac{H_k^2}{k}(-1)^k$$

Let $k+n = i$

$$\sum_{k=1}^\infty\sum_{i=k}^\infty\frac{H_{i}^2}{i}(-1)^{i}-\sum_{k=1}^\infty\frac{H_k^2}{k}(-1)^k=\sum_{i=1}^\infty\sum_{k=1}^i\frac{H_{i}^2}{i}(-1)^{i}-\sum_{k=1}^\infty\frac{H_k^2}{k}(-1)^k$$

$$S=\sum_{k=1}^\infty H_{k}^2(-1)^{k}-\sum_{k=1}^\infty\frac{H_k^2}{k}(-1)^k$$

The first diverges and the second converges hence I am tempted to say it diverges.

Question

  1. Is my method correct? I am concerened about interchanging the sums.
  2. I am not concerned about evaluating the series as much as seeing the limiting behaviour of the partial sums, when I put it in W|A it says it diverges by the limit test but i cant trust that or can I ?

1 Answers 1

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Let we consider the restricted sum $$ T(N)=\sum_{\substack{1\leq a \leq N \\ 1\leq b \leq N \\ a+b\leq N}}\frac{H_{a+b}^2}{a+b}(-1)^{a+b} \tag{1}$$ By considering $a+b=m$ as a new summation variable, we get that: $$ T(N) = \sum_{m=2}^{N}(-1)^m \frac{H_m^2}{m}\sum_{a=1}^{m-1}1 = \sum_{m=2}^{N}(-1)^m H_m^2\left(1-\frac{1}{m}\right) \tag{2}$$ and $\lim_{N\to +\infty}T(N)$ does not exist, since the general term of the last series behaves like $\log^2(m)$ in absolute value, making $\sum_{m=2}^{+\infty}(-1)^m H_m^2 $ a non-convergent series.

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    The first notation seems strange for me. Is it used interchangeably with the one I posted ? – 2017-02-22
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    @ZaidAlyafeai: you considered a double series without knowing if it was convergent or if we were allowed to switch the sums. I just considered the partial sums "over large triangles" delimited by $1\leq a,b\leq N$ and $a+b\leq N$. They lead to a non-convergent series. You may use this result to prove that the partial sums "over squares" delimited by $1\leq a,b\leq N$ lead to a non convergent-series too, so we are doomed. – 2017-02-22