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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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    @did 71 minutes until what?2012-12-09

3 Answers 3

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$ \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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    very nice solution2014-09-09
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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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    @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!2012-12-09
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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 $1/e^{n/2}$.