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Does the series

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

converge? I was trying to use this as a convergent majorant for proving convergence of $$\sum_{n=1}^\infty (-1)^n \frac {\ln(n)} n $$ but I'm not sure if that would work. Any help is appreciated!

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    $\begingroup$ It ($\sum\limits_{n=1}^{\infty}(-1)^n$) doesn't converge since $(-1)^n$ does not go to $0$. Remember that $\sum\limits_{n=1}^{\infty}a_n$ can only converge if $\lim\limits_{n\to\infty}a_n=0$. That series is an interesting series called Grandi's series though, and you can find more information about it at its Wikipedia page. $\endgroup$ – Minus One-Twelfth Apr 7 '19 at 11:59
  • $\begingroup$ Suppose it did. What do you imagine it would convergent to? $\endgroup$ – MJD Apr 7 '19 at 11:59
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    $\begingroup$ Second: You can prove convergence using a majorant, only if the proposed majorant is positive. $\endgroup$ – GEdgar Apr 7 '19 at 11:59
  • $\begingroup$ @MJD I figured maybe 1 or $ 0$, but I'm not really convinced of it (hence my question here). $\endgroup$ – Benjamin Caris Apr 7 '19 at 12:02
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    $\begingroup$ @KaviRamaMurthy Did not. However, I see what you mean now: it is alternating between -1 and 0, so indeed it does not converge. $\endgroup$ – Benjamin Caris Apr 7 '19 at 12:05
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Think about what $\sum_{n=1}^{\infty}(-1)^{n}$ means

You get $-1 + 1 -1 + 1...$

Does that converge to a number? Clearly not, it goes back and forth between $-1$ and $0$

A convergent sum will get closer and closer to one particular value, and can in fact get arbitrarily close to that value, which clearly our sum does not.

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Given $n\geq 1$, you have that $\frac{\ln n}{n}\geq 0$. Moreover, $n\mapsto \frac{\ln n}{n}$ is a decreasing function, which tends to $0$ as $n\to\infty$. Thus, you may apply the Alternating Series Test to conclude that the series $\sum_{n=1}^{\infty}(-1)^{n}\frac{\ln n}{n}$ converges.

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