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I would like to know how I proof that the following infinite series is divergent: $$\sum_{i=1}^n (-1)^n*(\frac{n}{n+1})^n$$ I tried the following tests: ratio test, root test, Leibniz criterion, divergence test, but they lead to an inconculsive result. The other only tests I am allowed to use is the comparison test and Cauchy's convergence test , but I don't know how to proof that the infinite series is divergent with these.

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  • $\begingroup$ Do you know $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$? Can you apply that to this series? $\endgroup$ – robjohn Oct 9 '16 at 0:56
  • $\begingroup$ $$\displaystyle {n \over n+1} = 1 - {1 \over n+1}$$ $\endgroup$ – Decaf-Math Oct 9 '16 at 1:03
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Observe \begin{align} \limsup_{n\rightarrow \infty}\ (-1)^n \left(\frac{n}{n+1}\right)^n = e^{-1} \neq 0. \end{align}

Edit: By the request of Ted Shifrin, I have updated my answer.

Heuristically speaking, an infinite series converges provided the terms that you are adding eventually becomes negligible. Otherwise, if you keep on adding ''big" terms to your partial sum \begin{align} \sum^N_{n=1} a_n + "\text{infinitely many big terms}" \end{align} then there's no hope that the series will converge. For instance consider the example \begin{align} -1 + 1 + -1 + 1 + -1+\ldots \end{align} We see that if we add an even number of terms then the partial sum will be $0$, but if we add an odd number of terms the partial sum will be $-1$. Hence our partial sum will fluctuate between $0$ and $-1$, which means the sequence of partial sums will not converge. The reason why the partial sum does not converge is mainly due to that fact that each term we add to the partial sum is not negligible. In fact the new term that we added to the partial sum greatly affects the value of the partial sum.

Back to the problem. In the given series, we see that for large enough $N$ the series looks more or less like the following \begin{align} \ldots + e^{-1}-e^{-1} + e^{-1}-e^{-1} + \ldots \end{align} which is almost exactly the same as our above example. Hence there's no way the given series converges.

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  • $\begingroup$ Why limit supremum? The limit of the series' sequence must be zero, not its limit supremum. $\endgroup$ – DonAntonio Oct 9 '16 at 0:39
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    $\begingroup$ @DonAntonio If the limit exists then it's equal to its limsup. $\endgroup$ – Jacky Chong Oct 9 '16 at 0:40
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    $\begingroup$ @Ja Yes...so what? Following the logic of what you wrote, the series $\;\sum a_n=-1+0-1+0-1+0-\ldots\;$ has a possibility to converge since $\;\lim\sup a_n=0\;$ ...though I think I can understand what you tried to convey. But stillI think it may be confusing. $\endgroup$ – DonAntonio Oct 9 '16 at 0:41
  • $\begingroup$ I wish people who answer questions here would think a bit more about pedagogy and the level of the OP. My interpretation, having read the question, is that this questioner won't have any idea what limsup is. So no need to bring it in. $\endgroup$ – Ted Shifrin Oct 9 '16 at 0:45
  • $\begingroup$ @DonAntonio My explanation is similar to the divergence test which states that if $\sum a_n$ converges then $a_n\rightarrow 0$. But just because $a_n\rightarrow 0$ doesn't mean $\sum a_n$ converges. $\endgroup$ – Jacky Chong Oct 9 '16 at 0:57
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The series has the $n^{th}$ term $a_n=\dfrac{1}{\left(-1-\dfrac{1}{n}\right)^n}$ does not converge to $0$. Thus the series diverges.

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  • $\begingroup$ Distributing the minus sign is the wrong thing to do and sets a bad example. But you omit the most important part. Presumably the OP did not think to try to find $\lim\limits_{n\to\infty} \left(\dfrac n{n+1}\right)^n$ or does not know how to do so. $\endgroup$ – Ted Shifrin Oct 9 '16 at 0:47
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    $\begingroup$ I don't agree with Ted in this one: distributing the sign makes it very clear what the series' general sequence is (and not "odd and even" elements and etc.), and then pointing out the sequences does not converge to zero wraps it all. If the OP didn't even think to find that limit, or even worse: doesn't know how to do it, then this exercise is probably beyond his capabilities right now, as proving the sequence converges to zero is one of the first, most basic things studied in infinite series...and, of course, limits must be a thoroughly studied and exercised subject by this time. +1 $\endgroup$ – DonAntonio Oct 9 '16 at 9:20

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