What is $\lim_{n\to\infty} n (\sum_{k = 1}^{n} \frac{1}{\sqrt {n^2 + k}} - 1)$? I am trying to find this limit: $\lim_{n \to \infty} a_n^{b_n}$, where $a_n = \sum_{k = 1}^{n} \frac{1}{\sqrt{n^2+ k}}$ and $b_n = n$. I have managed to prove that $\lim_{n \to \infty} a_n = 1$, by using the squeeze lemma (bounding $a_n$ between $1$ and $\frac{n}{\sqrt {n^2 + n}}$) and I want to use this theorem: if $\lim_{n \to \infty} = 0$, then $\lim_{n\to\infty} (1 + a_n) ^ {\frac{1}{a_n}} = e$. To apply this theorem, I did:
$$\lim_{n\to\infty}a_n^{b_n} = \lim_{n\to\infty}((1 + a_n - 1) ^ \frac{1}{a_n - 1}) ^ {b_n(a_n - 1)} = e^{\lim_{n\to\infty}b_n(a_n - 1)}$$. Now I am having problems finding the limit in the exponent. Could you help?
 A: We can rewrite the limit as
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{\sqrt{n^2+k}}-n = \lim_{n\to\infty}\sum_{k=1}^n \frac{n-\sqrt{n^2+k}}{\sqrt{n^2+k}} = \lim_{n\to\infty}\sum_{k=1}^n \frac{1-\sqrt{1+\frac{k}{n^2}}}{\sqrt{1+\frac{k}{n^2}}}$$
Since the numerator approaches zero we will rationalize and continue to simplify
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{-\frac{k}{n^2}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{n^2}} = \lim_{n\to\infty}\sum_{k=1}^n \frac{-\frac{k}{n}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{n^2}}\cdot\frac{1}{n}$$
We can sandwich this complicated expression with two other limits
$$\lim_{n\to\infty}-\frac{1}{2}\sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n} < \lim_{n\to\infty}\sum_{k=1}^n \frac{-\frac{k}{n}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{n^2}}\cdot\frac{1}{n} < \lim_{n\to\infty}\frac{-1}{\sqrt{1+\frac{1}{n}}+1+\frac{1}{n}}\sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n}$$
Both of which approach the Riemann sum
$$\longrightarrow -\frac{1}{2}\int_0^1 x\:dx = -\frac{1}{4}$$
Thus the limit is $\boxed{-\frac{1}{4}}$ by squeeze theorem.
A: We have
$$n\left(\sum_{k=1}^n{1\over\sqrt{n^2+k}}-1 \right)=n\sum_{k=1}^n\left({1\over\sqrt{n^2+k}}-{1\over n} \right)=n\sum_{k=1}^n{n-\sqrt{n^2+k}\over n\sqrt{n^2+k}}\\=\sum_{k=1}^n{-k\over\sqrt{n^2+k}\left(n+\sqrt{n^2+k}\right)}$$
and, for $1\le k\le n$,
$${1\over(n+1)(2n+1)}\le{1\over\sqrt{n^2+k}\left(n+\sqrt{n^2+k}\right)}\le{1\over n(2n)}$$
while $\sum_{k=1}^nk=n(n+1)/2$. Thus (momentarily dropping the minus sign in the numerator $-k$) we have
$${n\over2(n+1)(2n+1)}\le\sum_{k=1}^n{k\over\sqrt{n^2+k}\left(n+\sqrt{n^2+k}\right)}\le{(n+1)\over4n}$$
so by the squeeze theorem (and putting the minus sign back in), we have
$$\lim_{n\to\infty}n\left(\sum_{k=1}^n{1\over\sqrt{n^2+k}}-1 \right)=-{1\over4}$$
