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$\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}(\sum_{k=1}^{k=n}\dfrac{1}{k})}{n}$

Is this series convergent?

If I let $a_n=\dfrac{(-1)^{n+1}(\sum_{k=1}^{k=n}\dfrac{1}{k})}{n}=\dfrac{(-1)^{n+1}(1+\dfrac{1}{2}+...+\dfrac{1}{n})}{n}$

and found $\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n}{n+1}\dfrac{(1+\dfrac{1}{2}+...+\dfrac{1}{n+1})}{(1+\dfrac{1}{2}+...+\dfrac{1}{n})}=\dfrac{n(1+\dfrac{1}{2}+...+\dfrac{1}{n})+\dfrac{n}{n+1}}{n(1+\dfrac{1}{2}+...+\dfrac{1}{n})+(1+\dfrac{1}{2}+...+\dfrac{1}{n})}$

and since $\dfrac{n}{n+1}<1<(1+\dfrac{1}{2}+...+\dfrac{1}{n})$


But I realised that this is insufficient to say that the series converges

since $lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ may still be 1 which then the ratio test is inconclusive.

Any method to determine the convergence or divergence of the series?

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Hint: The series is alternating. What more would you need to show it converges? – Harald Hanche-Olsen Sep 8 '13 at 16:46
If we omit the minus signs, one can show that the series diverges. To prove your result, you need to show that the absolute values of the terms are decreasing, and have limit $0$. – André Nicolas Sep 8 '13 at 16:54
up vote 1 down vote accepted

Recall that a series $\sum a_n$ is called an Alternating Series if (i) the $a_n$ alternate in sign and (ii) $|a_{n+1}|\le |a_n|$ for all $n$ and (iii) $\lim_{n\to\infty} a_n=0$. Any alternating series converges.

To show (ii) in our case, we need to show that $$\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}}{n+1} \le \frac{1+\frac{1}{2}+\cdots +\frac{1}{n}}{n}.$$ After "cross-multiplication" and the obvious cancellation, this comes down to showing that $$\frac{n}{n+1}\le 1+\frac{1}{2}+\cdots +\frac{1}{n},$$ which is clear.

To prove (iii), maybe use the fact that $$1+\frac{1}{2}+\cdots+\frac{1}{n}\lt \int_1^{n+1}\frac{dx}{x}=\log(n+1)$$ and $\lim_{n\to\infty}\frac{\log(n+1)}{n}=0$.

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Is it really true that decreasing absolute values and terms going to zero is built into the definition of “alternating series”? I find that disturbing somehow. It seems to me that only the alternating signs belong in the definition. – Harald Hanche-Olsen Sep 8 '13 at 18:59
It is built into the definition in all the calculus books I have used as texts. One has to keep reminding students that "alternating series" means more than signs bounce! – André Nicolas Sep 8 '13 at 19:17

The series is convergent. We can indeed use a summation by parts argument. We have, denoting $s_k:=\sum_{i=0}^k(-1)^i$ \begin{align*} \sum_{n=M}^{M+N}(-1)^n\frac{\sum_{j=1}^n\frac 1j}{n}&=\sum_{l=M}^{M+N}s_l\frac{\sum_{j=1}^l\frac 1j}{l}-\sum_{l=M-1}^{M+N-1}s_l\frac{\sum_{j=1}^{l+1}\frac 1j}{l+1}, \end{align*} hence $$\left|\sum_{n=M}^{M+N}(-1)^n\frac{\sum_{j=1}^n\frac 1j}{n}\right|\leqslant \frac{2\log(M+N)}{M+N}+\frac{2\log M}M+\sum_{l=M}^{M+N-1}\left|\frac{\sum_{j=1}^{l+1}\frac 1j}{l+1}-\frac{\sum_{j=1}^{l}\frac 1j}{l}\right|.$$ Since $$\left|\frac{\sum_{j=1}^{l+1}\frac 1j}{l+1}-\frac{\sum_{j=1}^{l}\frac 1j}{l}\right|\leqslant 2\log l\left(\frac 1l-\frac 1{l+1}\right)+\frac 1{(l+1)^2}$$ and the series $\sum_{k=1}^\infty\frac{\log k}{k^2}$ is convergent, we are done.

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Note that $$\lim_{n\to\infty}\frac{\sum_{k=1}^{k=n}\dfrac{1}{k}}{n}=\lim_{n\to\infty}\frac{1/n}{n+1-n}=0,$$ and that $$\frac{\sum_{k=1}^{k=n}\dfrac{1}{k}}{n}-\frac{\sum_{k=1}^{k=n+1}\dfrac{1}{k}}{n+1}=\frac{1}{n(n+1)}\sum_{k=1}^{k=n}\dfrac{1}{k}-\frac{1}{(n+1)^2}>0,$$ for all $n$. So $\frac{\sum_{k=1}^{k=n}\dfrac{1}{k}}{n}$ decrease to $0$. By "alternating series test" we know the series converge.

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