<s>[INST] 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?
 [/INST] 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.
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