What is $\lim_{n\to\infty} n (\sum_{k = 1}^{n} \frac{1}{\sqrt {n^2 + k}} - 1)$?

Solution 1:

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.

Solution 2:

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

$$\lim_{n\to\infty}n\left(\sum_{k=1}^n{1\over\sqrt{n^2+k}}-1 \right)=-{1\over4}$$