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I am trying to decide if $$\sum_{n=1}^\infty \left(\frac{1}{n}-1\right)^{n^2}$$ converges. By the alternating series test, as far as I can see, the series converges. This is also true by the root test. In both cases I assume that $$\left|\left(\frac{1}{n}-1\right)^{n^2}\right| = \left(1-\frac{1}{n}\right)^{n^2}$$ and I can't see anything wrong with that. What makes me unsure is that Wolfram Alpha says that the sum does not converge.

Wolfram Alpha uses the limit test, which I am not able to complete: $$\lim_{n\to\infty} a_n = \lim_{n\to\infty}\left(\frac{1}{n}-1\right)^{n^2} = \lim_{n\to\infty} e^{n^2 \ln{\frac{1-n}{n}}}$$ The problem is the logarithm which is not defined for any $n$, and I am not aware of another way to compute that limit.

So does this series converge?

Edit: When I ask Wolfram Alpha to evaluate the sum, it says that the seires do not converge. But it also says that "computation timed out". Maybe this could be the reason for why it's wrong.

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  • $\begingroup$ Re Wolfram Alpha, you could have asked the sum of $(-1)^n(1-1/n)^{n^2}$, it works all right and gives the numerical value $0.04376073746559...$ $\endgroup$
    – Did
    Mar 22, 2011 at 9:42
  • $\begingroup$ Didier: Good idea. But then, if $(-1)^n(1-1/n)^{n^2} = (1/n-1)^{n^2}$ (which seems clear to me), Wolfram Alpha is inconsistent. $\endgroup$
    – Eivind
    Mar 22, 2011 at 9:51
  • $\begingroup$ Oops, sorry. $\endgroup$
    – anonymous
    Mar 22, 2011 at 9:59
  • $\begingroup$ @Elvind: Yes. I am not an expert in WA behaviour but I seem to remember having read about other similar inconsistencies before. This simply means that WA is a tool, imperfect like many tools and useful like some of them. $\endgroup$
    – Did
    Mar 22, 2011 at 11:46
  • $\begingroup$ I have now also tried in $Mathematica$, and it has the same problem. But even though it says that the series doesn't converge, it gives me a numerical answer when I ask for it. $\endgroup$
    – Eivind
    Mar 22, 2011 at 14:47

2 Answers 2

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$$\left| \left( \frac{1}{n} - 1\right)^{n^2} \right| = \left| \left( \frac{1}{n} - 1\right) \right|^{n^2} = \left( 1 - \frac{1}{n}\right)^{n^2}$$ as you correctly noted. And we have $$\limsup_{n\to\infty} \sqrt[n]{\left( 1 - \frac{1}{n}\right)^{n^2}} = \limsup_{n\to\infty} \left( 1 - \frac{1}{n}\right)^n = \frac{1}{e} < 1$$ so this series converges.

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  • $\begingroup$ That is the same conclusion I reached. Is Wolfram Alpha wrong then? Also, I am not familiar with $\text{limsup}$. What does that mean? $\endgroup$
    – Eivind
    Mar 22, 2011 at 9:37
  • $\begingroup$ Eivind: $\limsup$ is the supremum of the partial limits, in a sense it is the highest number that a sequence has a converging subsequence. If the sequence converges as a whole, then $\limsup = \lim$. $\endgroup$
    – Asaf Karagila
    Mar 22, 2011 at 9:39
  • $\begingroup$ Asaf: OK. Is it necessary to use it in this case? Why not a regular limit? $\endgroup$
    – Eivind
    Mar 22, 2011 at 9:44
  • $\begingroup$ Then your conclusion is correct. $\limsup_{n\to\infty} a_n := \inf_{n\in\mathbb N}\sup_{k\geq n} a_k$, that is the $\sup \{a \; |\; a \text{ is a limit point of } (a_n)\}$. $\endgroup$ Mar 22, 2011 at 9:44
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    $\begingroup$ Note that a limit does not have to exist for $\limsup$ to exist, therefore the $\limsup$-formulation of the root test is somewhat more general (stronger). In the case we have considered a regular limit would suffice to show the series convergent. $\endgroup$ Mar 22, 2011 at 9:51
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Hint: Consider $(1-1/n)^n \leq e^{-1}$.

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