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Prove that the following sequence is convergent and calculate it's limit:

$ a_n = (1+\frac{1}{n^2+2n} ) ^{n^2} $

I had this question in an exam today and failed miserably. I tried using squeeze theorem, sub-sequences, but I could not find a way. Intuitively the limit is e, how to solve this? Thanks in advance.

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  • $\begingroup$ This should make one think of the exponential map. $\endgroup$ – J.F Jan 19 '19 at 19:51
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An idea:

$$\left(1+\frac1{n^2+2n}\right)^{n^2}=\left(1+\frac1{n^2+2n}\right)^{n^2+2n}\left(1+\frac1{n^2+2n}\right)^{-2n}\xrightarrow[n\to\infty]{}e\cdot1=e$$

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  • $\begingroup$ $(1+\frac{1}{n^2+2n} ) ^{-2n} $ converges into 1 by using squeeze theorem, right? $\endgroup$ – Tegernako Jan 20 '19 at 7:15
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    $\begingroup$ @Tegernako Certainly:$$\left(1+\frac1{n^2+2n}\right)^{-2n}=\frac1{\left(1+\frac1{n^2+2n}\right)^{2n}}=\left(\frac{(n+1)^2}{n^2+2n=(n+1)^2-1}\right)^{2n}$$and $$1=\left(\frac{(n+1)^2}{(n+1)^2}\right)^{2n}\le\left(\frac{(n+1)^2}{(n+1)^2-1}\right)^{2n}\le\frac{(n+1)^2}{(n+1)^2-1}\xrightarrow[n\to\infty]{}1$$ $\endgroup$ – DonAntonio Jan 20 '19 at 13:29
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$$\ln a_n=n^2\ln\left(1+\frac1{n^2+n}\right)=n^2\left(\frac1{n^2+n} +O(n^{-4})\right)=\frac{n^2}{n^2+2}+O(n^{-2})$$ etc.

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