# Proving the limit of the following sequence

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.

• This should make one think of the exponential map. – J.F Jan 19 at 19:51

$$\left(1+\frac1{n^2+2n}\right)^{n^2}=\left(1+\frac1{n^2+2n}\right)^{n^2+2n}\left(1+\frac1{n^2+2n}\right)^{-2n}\xrightarrow[n\to\infty]{}e\cdot1=e$$
• $(1+\frac{1}{n^2+2n} ) ^{-2n}$ converges into 1 by using squeeze theorem, right? – Tegernako Jan 20 at 7:15
• @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 $$1=\left(\frac{(n+1)^2}{(n+1)^2}\right)^{2n}\le\left(\frac{(n+1)^2}{(n+1)^2-1}\right)^{2n}\le\frac{(n+1)^2}{(n+1)^2-1}\xrightarrow[n\to\infty]{}1$$ – DonAntonio Jan 20 at 13:29
$$\ln a_n=n^2\ln\left(1+\frac1{n^2+n}\right)=n^2\left(\frac1{n^2+n} +O(n^{-4})\right)=\frac{n^2}{n^2+2}+O(n^{-2})$$ etc.
When finding the limit, the $$2n$$ will drop out of $$n^2+2n$$. Thus, the limit is the limit as $$n$$ approaches infinity in $$(1+\frac{1}{n^2} ) ^{n^2}$$. Substituting $$m$$ for $$n^2$$ gives the limit as $$m$$ approaches infinity of $$(1+\frac{1}{m} ) ^{m}$$, which is equal to $$e$$. Note that $$n^2$$ can be replaced with $$m$$ because they both approach infinity.