Test $f_n = (1+\frac{1}{n^2})^n$ for convergence and give its limit if it exists. Now this exercise looks like I need to use $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e$$ somehow, but I'm unsure of how. Bernoulli would give me
$$\left(1+\frac{1}{n^2}\right)^n > 1+n\cdot\frac{1}{n^2}=1+\frac{1}{n}$$ for which the limit is $1$, but I'm not sure how that helps. Can anybody give me a tip on how to solve this elegantly?
 A: Bernoulli's inequality is a very powerful tool as seen below.

We have via Bernoulli's inequality for $n>1$ $$1-\frac{1}{n^3}\leq \left(1-\frac{1}{n^4}\right)^{n}\leq 1\tag{1}$$ and hence by Squeeze Theorem we have $$\lim_{n\to\infty} \left(1-\frac{1}{n^4}\right)^{n}=1\tag{2}$$ In exactly the same manner we can show that $$\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^{n}=1\tag{3}$$ Dividing equation $(2)$ by equation $(3)$ we get $$\lim_{n\to\infty} \left(1+\frac{1}{n^{2}}\right)^{n}=1\tag{4}$$ The same technique can be used to prove the general result that $(1\pm n^{p}) ^{n} \to 1$ as $n\to\infty$ if $p<-1$ and one observes that the above makes no use of symbols like $e, \exp, \log$. 
A: $$\ln f_n=n\ln\left(1+\frac1{n^2}\right)=n\left(\frac1{n^2}+O(n^{-4})\right)=\frac1n+O(n^{-3})\to0$$
as $n\to\infty$. So $f_n\to1$.
A: You have
$$
f_n=\exp\left[n\log\left (1+\frac1 {n^2}\right)\right]= \exp\left[\frac 1n+o \left(\frac1 {n^3}\right)\right]\to e^0=1
$$
as $n \to \infty$.
A: $$\left(1+\frac{1}{n^2}\right)^n\geq 1+\frac{1}{n}\quad\text{by Bernoulli's inequality or the binomial theorem}$$
$$ \left(1+\frac{1}{n^2}\right)^n\leq \exp\frac{n}{n^2}\leq \frac{1}{1-\frac{1}{n}}\quad\text{by }e^x\geq x+1\text{ and }e^x\leq\frac{1}{1-x}\text{ for }x\in(0,1) $$
so the limit is $1$ by squeezing.
A: The expression equals
$$\left [\left(1+\frac{1}{n^2}\right)^{n^2}\right]^{1/n}$$
Inside the brackets the limit is $e.$ Thus for large $n,$
$$2 < \left(1+\frac{1}{n^2}\right)^{n^2}<3.$$
For such $n,$
$$2^{1/n} < \left[\left(1+\frac{1}{n^2}\right)^{n^2}\right]^{1/n}<3^{1/n}.$$ 
Since the left side and the right side both $\to 1,$ our limit is $1$ by the squeeze theorem.
A: A proof unaware of the exponential/logarithm function or the constant $e$. On the one hand, $1≤ 1+1/{n^2}$ and hence $1≤(1+\frac{1}{n^2})^{n}$. On the other hand, by the binomial theorem,
$$\left(1+\frac{1}{n^2}\right)^{n} = \sum_{k=0}^n \binom{n}{k}n^{-2k}$$
The first term in this series is $1$; so lets try to prove that the remainder of the series is $O(1/n)$. This is
$$ \sum_{k=1}^n \binom{n}{k} n^{-2k} = \sum_{k=0}^{n-1}\binom{n}{k+1} n^{-2k-2} = \frac{1}{n^2}\sum_{k=0}^{n-1} \binom{n}{k+1}n^{-2k}$$
Now use the crude inequality $\binom{n}{\ell} ≤ n^\ell$ to get that
$$ \sum_{k=1}^n \binom{n}{k} n^{-2k} ≤ \frac{1}{n^2}\sum_{k=0}^{n-1} {n^{k+1}}n^{-2k} = \frac1n\sum_{k=0}^{n-1} {n^{-k}} $$
At this point we can e.g. claim that $n≥2$ so that $n^{-k} ≤ 2^{-k}$ and add countably many terms to the sum to see that
$$\sum_{k=0}^{\infty} {n^{-k}} ≤ \sum_{k=0}^{\infty} {2^{-k}} < 2 <\infty$$
Therefore we have proved that:
$$ 1 ≤ \left(1+\frac{1}{n^2}\right)^{n} ≤ 1 + \frac{2}{n}\to 1$$
so by the squeeze rule, the limit is $1$.
