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Investigate the convergence of the two series

  1. $$\sum_{n=1}^\infty \frac{2}{n^n}$$
  2. $$\sum_{n=1}^\infty (-1)^{n-1} \frac{2^n}{n^2}$$

Attempt

  1. $$\frac{u_{n+1}}{u_n}=\frac{n^n}{(n+1)^{n+1}}=\frac{1}{n(1+1/n)^{n+1}}\to0<1$$ then by D' Alemberts' test the series is convergent. Correct?

  2. Let $v_n= \frac{2^n}{n^2}$ I want to check the convergence by Leibnitz's test. How to show that $\{v_n\}$ is monotonic decreasing and $v_n\to 0$ as $n\to \infty$

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  • 2
    $\begingroup$ Note that $2^n/n^2\to+\infty$ $\endgroup$ – Robert Z May 14 '18 at 17:43
  • $\begingroup$ @RobertZ Means it is divergent. Q1. How to check $2^n/n^2\to+\infty$, Q2. Is it sufficient that $2^n/n^2\to+\infty$ implies series is divergent? $\endgroup$ – user1942348 May 14 '18 at 17:54
  • $\begingroup$ But D' Alembert's theorem is applicable for the series of positive terms? Here it is an alternating series. $\endgroup$ – user1942348 May 14 '18 at 17:58
  • $\begingroup$ Sorry. I mean $\frac{|u_{n+1}|}{|u_n|}\to 2>1$ implies that $|u_n|\to +\infty$ and therefore the series is not convergent (a necessary condition for convergence is $u_n\to 0$). $\endgroup$ – Robert Z May 14 '18 at 18:03
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    $\begingroup$ @user1942348 If the general terms of a series do not approach $0$, then the series diverges. Inasmuch as $\frac{(-1)^{n-1}2^n}{n^2}$ does not approach $0$, the series diverges. That is all we need. $\endgroup$ – Mark Viola May 14 '18 at 18:25
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Your answer to (1) is correct.

For (2), the ratio is $\dfrac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^{n}}{n^2}} =\dfrac{2}{(1+1/n)^2} \to 2 $ so this sum diverges.

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Note that for $v_n= \frac{2^n}{n^2}$, both conditions do not hold: $$1) \ v_{n+1}>v_n \iff \frac{2^{n+1}}{(n+1)^2}>\frac{2^n}{n^2} \iff 2n^2>(n+1)^2 \iff n^2>2n+1,n>2\\ 2) \ \lim_{n\to\infty} v_n=\lim_{n\to\infty} \frac{2^n}{n^2}\overbrace{=}^{L'H}\lim_{n\to\infty} \frac{2^n\ln 2}{2n}\overbrace{=}^{L'H} \lim_{n\to\infty} \frac{2^n\ln^22}{2}=\infty.$$

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