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Can someone please help me prove that this series is convergent? $$ \sum_{i=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2} $$

I guess I'm supposed to show that the limit of the sequence is an "e" limit, that means something of the kind: $$ \left( 1\pm{1 \over a} \right)^a $$ But how? I came to this state by now and that's where I'm stuck:

$$ \left( {n^2+n+1-n \over {n^2+n+1}} \right)^{n^2} = \left( 1+{n \over {n^2+n+1}} \right)^{n^2} $$

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Sorry, but my answer wasn't very good. – Thomas Dec 9 '12 at 21:15
Thank you for your help anyway. – Squidward Dec 9 '12 at 21:16
71 minutes. $ $ – Did Dec 9 '12 at 22:19
@did 71 minutes until what? – apnorton Dec 9 '12 at 22:25
up vote 3 down vote accepted

The $n$th term is $\left(1-\dfrac{n}{n^2+n+1}\right)^{n^2}<\left(\left(1-\dfrac{1}{2n}\right)^n\right)^n$. I would suggest a limit comparison test with $1/e^{n/2}$.

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+1 Nice answer. – Thomas Dec 9 '12 at 21:54

Note that $${\left( \dfrac{n^2+n+1}{n^2+1}\right)^{n^2}}={\left(1+ \dfrac{n}{n^2+1}\right)^{\frac{n^2+1}{n}\cdot\frac{n^3}{n^2+1}}}\geqslant 2^{\frac{n}{2}},$$ because $2<\left(1+ \dfrac{n}{n^2+1}\right)^{\frac{n^2+1}{n}}<3$ and $\frac{n^3}{n^2+1}\geqslant \frac{n}{2}$ for $n\geqslant 1.$ Therefore, $$\left( \dfrac{n^2+1}{n^2+n+1}\right)^{n^2}\leqslant {\frac{1}{2^{\frac{n}{2}}}}=\left(\frac{1}{\sqrt{2}}\right)^n$$ and the series $$\sum_{n=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2}$$ converges by comparison test.

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Thanks, it's quite overwhelming the way you've got the answer, but @Jonas Meyer 's way is much more straightforward. – Squidward Dec 9 '12 at 21:54
@Squidward, it is important to understand this approach too as it is helpful to have tricks and tools to throw at problems. These help your imagination and you never know when trying different things leads to prosperous results! – Amzoti Dec 9 '12 at 22:24

$$ \left(1-\dfrac{n}{n^2+n+1}\right)^{n^2}\lt\exp\left(-\dfrac{n}{n^2+n+1}\cdot n^2\right)\lt\mathrm e^{-n+1} $$

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This is a very interesting solution, which took me some time to figure out why it's true. I was wondering what made you think about it, what was the intuition? – user1685224 May 1 '13 at 18:38
@user1685224 The fact that $1-x\leqslant\mathrm e^{-x}$ for every $x$ is useful in many different contexts, in fact as soon as some products are involved, because the exponential function is well suited to products (and powers are nothing but specific products). – Did May 1 '13 at 19:30
Thank you very much! Learned a new thing. – user1685224 May 3 '13 at 4:44
very nice solution – avz2611 Sep 9 '14 at 17:45

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