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Does $\displaystyle\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{(-1)^{n+k}}{n}$ diverge?

It is clear that the alternating Harmonic series converges: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\log 2.$$ Thus, $S_k=\displaystyle\sum_{n=k}^\infty \frac{(-1)^{n+k}}{n}$ converges for each $k$. For each $k$, the sum could be expressed either as $\log 2-\alpha$ or $\alpha -\log 2$. So, we're really only interested in how much $S_k$ deviates $(\alpha)$ from the alternating series $S_1$. The numerators of $S_k$ follow for even $k$ and for odd $k$.

However, it seems that the partial sums may slowly go towards infinity as shown in this Wolfram plot here. Reasonably, since the difference between $S_1$ and $S_k$ probably behaves like $O(1/k)$, the sum probably diverges.

What would be the best way to show convergence/divergence?

Martin Sleziak
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user519208
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2 Answers2

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The sum of the first two terms is $$ \begin{align} &\phantom{=1}1-\frac12+\frac13-\frac14+\frac15-\frac16+\dots\\ &\phantom{=1-1}\frac12-\frac13+\frac14-\frac15+\frac16-\dots\\ &=1 \end{align} $$ The sum of the next two terms is $$ \begin{align} &\phantom{=\frac13}\frac13-\frac14+\frac15-\frac16+\dots\\ &\phantom{=\frac13-\frac13}\frac14-\frac15+\frac16-\dots\\ &\phantom{1}=\frac13 \end{align} $$ The sum of the first $2n$ terms is $$ H_{2n}-\frac12H_n\sim\frac12\log(n)+\log(2)+\frac12\gamma $$ Thus, the series does diverge.

robjohn
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Let $A_k=\sum_{n=k}^{\infty}\frac {(-1)^{n+k}}{n+k}.$ We have $$A_k=\sum_{m=0}^{\infty}B(m,k)$$ where $B(m,k) =\frac {1}{2k+2m}-\frac {1}{2k+2m+1}.$

We have $B(m,k)>B(m+1,k)>0 $. So $$A_k>\sum_{m=0}^{k-1}B(m,k)\geq$$ $$\geq kB(k-1,k)=$$ $$=k(\frac {1}{4k-2}-\frac {1}{4k-1})=$$ $$=\frac {k}{(4k-2)(4k-1)}>$$ $$>\frac {k}{(4k)(4k)}=\frac {1}{16k}.$$ So $\sum_{k=1}^{\infty}A_k\geq \sum_{k=1}^{\infty}\frac {1}{16 k}=\infty.$

Note: $\frac {1}{16 k }$ is not meant to be an accurate estimate for $A_k.$ Just a useful lower bound.

DanielWainfleet
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