I need help with a limit problem I need help with the following limit:
$\lim_{n \to \infty}\left [ (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right ) \right ],a\in \mathbb{N}$
I am completely lost.
 A: Use the binomial theorem. 
$$\left(1 + \frac{2}{n^2}\right)^a - 1 = \sum_{k=1}^a \binom{a}{k} \frac{2^k}{n^{2k}}.$$
Multiplying by $n^2 + 1$ gives you
$$\sum_{k=1}^a\binom{a}{k}2^k\left(\frac{1}{n^{2k-2}} + \frac{1}{n^{2k}}\right).$$
Since $1/n^s \to 0$ iff $s > 0$, and since $2k > 0$ for all $k \ge 1$, we see that the only surviving term is $2k-2$ for $k=1$. In this case, the limit is $$\binom{a}{1} 2^1 = 2a.$$
A: Take: $t=\frac{1}{n^2}$
$$\lim_{n \to \infty}\left [ (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right ) \right ]=\lim_{t \to 0} \frac{(1+t)\left ( \left ( 1+2t \right )^a - 1 \right )}{t}=0/0$$
apply L-Hopital rule:
$$=\lim_{t \to 0}\{\left ( 1+2t \right )^a -1+2a(1+t)\left ( 1+2t \right )^{a-1}\}=2a$$
A: Let us consider $$A= (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right )$$ In a first time, let us focus on $(1+x)^a$ and use the generalized binomial theorem for generating a Taylor-like series when $x$ is small. So $$(1+x)^a=1+a x+\frac{1}{2} (a-1) a x^2+\frac{1}{6} (a-2) (a-1) a x^3+O\left(x^4\right)$$ Now, replace $x$ by $\frac{2}{n^2}$ $$\left ( 1+\frac{2}{n^2} \right )^a - 1 =\frac{2 a}{n^2}+\frac{2 (a-1) a}{n^4}+\frac{4 (a-2) (a-1) a}{3 n^6}+\cdots$$ Multiply by $(n^2+1)$, expand and simplify; we arrive to $$A=2 a+\frac{2 a^2}{n^2}+\frac{2 \left(2 a^3-3 a^2+a\right)}{3
   n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ which shows the limit and how it is approached.
