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$$\sum_1^\infty \frac{(-1)^n (1 + 1/2 + \cdots+ 1/n)}n$$

I tried applying the alternating series test, but I think that fails. I don't know which other test I could use here.

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  • $\begingroup$ Are you familiar with harmonic numbers ? $\endgroup$ – Claude Leibovici Nov 29 '14 at 6:44
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    $\begingroup$ Note that $(1+1/2+\cdots+1/n)\le 1+\ln n $; see here. $\endgroup$ – vadim123 Nov 29 '14 at 6:44
  • $\begingroup$ @ClaudeLeibovici I know that the harmonic series diverges. $\endgroup$ – bakhtiar Nov 29 '14 at 6:45
  • $\begingroup$ But your expression is $\sum_{n=1}^{\infty}\frac{(-1)^n H_n}{n}$ which makes a big difference. Vadim's comment can be very useful. $\endgroup$ – Claude Leibovici Nov 29 '14 at 6:50
  • $\begingroup$ The alternating series test works, but it takes some calculatio to show the conditions are met. $\endgroup$ – André Nicolas Nov 29 '14 at 6:51
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Let $b_n=\displaystyle\frac{1 + 1/2 + \cdots+ 1/n}n$. Then the series we want to consider is: $$\sum_{n=1}^\infty (-1)^bb_n.$$ As vadim123 said in the above comment, $1 + 1/2 + \cdots+ 1/n\leq 1+\ln n$, which implies that $$0<b_n\leq \frac{1+\ln n}{n}$$ which implies that $\lim_{n\to\infty} b_n=0$. On the other hand, we have $$(n+1)b_{n+1}=1 + 1/2 + \cdots+ 1/n+1/(n+1)=nb_n+1/(n+1),$$ which implies that $$n(b_{n+1}-b_n)=1/(n+1)-b_{n+1}<0.$$ That is, $b_n$ is decreasing. Therefore, by alternating series test, $\displaystyle\sum_{n=1}^\infty (-1)^bb_n$ converges.

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  • $\begingroup$ Thanks for your answer, Paul. I don't understand how does $0 < b_n \le \frac{1 + \ln n}{n}$ implies limit $b_n = 0$? $\endgroup$ – bakhtiar Nov 29 '14 at 7:00
  • $\begingroup$ Since $\lim_{n\to\infty}\frac{1+\ln n}{n}=0$, by squeezing theorem, we have $\lim_{n\to\infty} b_n=0$. $\endgroup$ – Paul Nov 29 '14 at 7:03
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An elementary version.

As you have already noticed, it remains to show, that $$ \frac{(1 + 1/2 + \cdots+ 1/n)}n \geq\frac{(1 + 1/2 + \cdots+ 1/n+1/(n+1))}{n+1}. $$ But it is equivalent to $$ (1 + 1/2 + \cdots+ 1/n)\left(\frac1{n(n+1)}\right)\geq\frac1{(n+1)^2}, $$ which is obvious.

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