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Does this series $$\sum\limits_{n=0}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}$$ converge?.

Attempt :

First I checked if $$u_n=\left(\dfrac{n}{n+1}\right)^{n^2}\underset{n\to +\infty}{\longrightarrow}0$$ because that is a necessary condition

$$u_n=\exp\bigg[n^2\bigg(\ln n-\ln(n+1)\bigg)\bigg]=\exp\bigg[-n^2\bigg(\ln (n+1)-\ln n\bigg)\bigg]$$

Equivalent of $\ln (n+1)-\ln n$ : According to the mean value theorem, $\exists c \in ]n,n+1[$ such that $\ln (n+1)-\ln n =f'(c)=\dfrac{1}{c}$ since $c\sim n$ so we have $\big(\ln(n+1)-\ln n\big) \sim \dfrac{1}{n} $ and $-n^2\bigg(\ln (n+1)-\ln n\bigg)\sim-n$

Thus :

$$\displaystyle \lim_{n\to +\infty} u_n =\lim_{n\to +\infty} e^{-n}=0$$

Now I need to find if this sequence is dominated by an another one, I already know it converges. But I am blocked...

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    $\begingroup$ Does $\sum_{n=1}^\infty e^{-n}$ converge? $\endgroup$ Oct 6 '17 at 17:05
  • $\begingroup$ @LordSharktheUnknown Yes, because $e^{-n}=o(n^{-2})$ $\endgroup$
    – Stu
    Oct 6 '17 at 17:07
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    $\begingroup$ That's an interesting argument! $\endgroup$ Oct 6 '17 at 17:08
  • $\begingroup$ Also, your difference of logs is just $\ln \left( \frac{n+1}{n}\right)$ which plainly goes to 0. $\endgroup$
    – Randall
    Oct 6 '17 at 17:08
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    $\begingroup$ @Stu While they're not equivalent, you can put exponential bounds on $u_n$ using that fact. In particular, note that $(\frac n{n+1})^n$ is less than $\frac12$ for all sufficiently large $n$, and so $(\frac n{n+1})^{n^2}$ is less than $2^{-n}$ for sufficiently large $n$. $\endgroup$ Oct 6 '17 at 17:29
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My own Idea Cauchy Rule $$a_n=\left(\dfrac{n}{n+1}\right)^{n^2}$$ $$\sqrt[n]{a_n}=\left(1+\dfrac{1}{n}\right)^{-n} =e^{-\frac{\ln(1+\frac{1}{n})}{\frac{1}{n}}} \overset{h=\frac{1}{n}}{=}e^{-\frac{\ln(1+h)}{h}}\to e^{-1}<1$$ Hence your series converges.

OP attempt By equivalence However, from OP Attempt, he should have shown instead that, $$\lim_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}\Longleftrightarrow a_n\sim e^{n} e^{-\frac{1}{2}}$$ and therefore, $$\sum\limits_{n=0}^{\infty}e^{-n}=\sum\limits_{n=0}^{\infty}(e^{-1})^{n}=\frac{1}{1-e^{-1}}$$ this gives another prove of the convergence.

Proof of $\lim\limits_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}$

Setting $h=\frac{1}{n}$

$$a_n =\left(1+\dfrac{1}{n}\right)^{-n^2} =e^{-\frac{\ln(1+\frac{1}{n})}{\frac{1}{n^2}}}{=}e^{-\frac{\ln(1+h)}{h^2}}$$ Thus,

$$a_ne^{n} = e^{\frac{1}{h}-\frac{\ln(1+h)}{h^2}}=e^{-\frac{1}{2}+o(h)}$$

Since $$\ln(1+h)= h -\frac{h^2}{2} +o(h^2) $$

Hence $$\lim_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}$$

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  • $\begingroup$ are you using this property : $\sum q^n$ converges if $|q|<1$ $\endgroup$
    – Stu
    Oct 6 '17 at 17:29
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    $\begingroup$ Not at all I am using this en.wikipedia.org/wiki/Root_test $\endgroup$
    – Guy Fsone
    Oct 6 '17 at 17:32
  • $\begingroup$ I haven't seen this theorem yet, that will come soon $\endgroup$
    – Stu
    Oct 6 '17 at 17:39
  • $\begingroup$ Ok check the edit the second will be surely satisfactory to you $\endgroup$
    – Guy Fsone
    Oct 6 '17 at 17:55
  • $\begingroup$ Perfect thank you very mush $\endgroup$
    – Stu
    Oct 6 '17 at 18:18
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Considering $$a_n=\left(\frac{n}{n+1}\right)^{n^2}\implies \log(a_n)=n^2\log\left(\frac{n}{n+1}\right)$$ $$\log\left(\frac{a_{n+1}}{a_n}\right)=(n+1)^2\log\left(\frac{n+1}{n+2}\right)-n^2\log\left(\frac{n}{n+1}\right)$$ $$\log\left(\frac{a_{n+1}}{a_n}\right)=(n+1)^2\log\left(1-\frac{1}{n+2}\right)-n^2\log\left(1-\frac{1}{n+1}\right)$$ COnsidering large $n$ and using Taylor series $$\log\left(\frac{a_{n+1}}{a_n}\right)=-1+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ and using $x=e^{\log(x)}$ $$\frac{a_{n+1}}{a_n}=\frac{1}{e}+\frac{1}{3 e n^2}+O\left(\frac{1}{n^3}\right)\sim \frac{1}{e}<1$$ then convergence

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We know that $$ \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{-n} = \lim_{n\to \infty}\left( \frac{n}{n+1}\right)^{n} = \frac{1}{e}<\frac{1}{2}. $$

Then, there is some $n_0$ such that $n\geq n_0$ implies that $$ \left(\frac{n}{n+1} \right)^{n} < \frac{2}{3} \implies \left(\frac{n}{n+1} \right)^{n^2} < \left(\frac{2}{3}\right)^{n}. $$

Therefore, your sum can be bounded by $$ \sum_{n\geq 1}\left(\frac{n}{n+1} \right)^{n^2} < \sum_{n< n_0}\left(\frac{n}{n+1} \right)^{n^2} + \sum_{n\geq n_0}\left(\frac{2}{3}\right)^{n} < \infty. $$

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I created a graph of all the partial sum values up to $41$, and it does seem to converge. In the ordered pair $(x, y)$ let $y \approx \sum_{n = 0}^x\big(\frac{n}{n + 1}\big)^{n^2}$ for $x = 1, 2, 3,\ldots$

$(1, 0.5)$

$(2, 0.697531)$

$(3, 0.772616)$

$(4, 0.800763)$

$(5, 0.811246)$

$(6, 0.815135)$

$(7, 0.816575)$

$(8, 0.817108)$

$(9, 0.817304)$

$(10, 0.817377)$

$(11, 0.817404)$

$(12, 0.817414)$

$\qquad \ \ \ \vdots$

$(40, 0.817419)$

$(41, 0.817419)$

As we can see, it converges. In fact: $$\sum_{n = 0}^{\infty}\bigg(\frac{n}{n + 1}\bigg)^{n^2} \approx 0.81741943329783346091309999437310691477482788227147414627301379318568020318099543837367156249468532191667154297066839$$

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  • $\begingroup$ This is only a conjecture. It is not sufficient to say that converges $\endgroup$
    – Stu
    Oct 7 '17 at 8:45

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