Evaluation of limit $\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right)$

Question

Evaluation of the limit $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right).$$

I don't understand how can I start the above problem because Riemann sums and integrals are not applicable here.

It seems that we can solve it using the Squeeze theorem, but I don't understand which inequality I have to use. Help required, thanks.

Answer 1

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

and, for $1\le k\le n$,

$${k\over\sqrt{1+{1\over n}}+1}\le{k\over\sqrt{1+{k\over n^2}}+1}\le{k\over 2}$$

so, since $\sum k=n(n+1)/2$,

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

Answer 2

Note that $\sqrt{1+k/n^2} -1 = \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right)$ by the Maclaurin series of $\sqrt{1+x}$. We use the notation $O(f(n))$ for some term thats bounded for sufficiently large $n$ by $C |f(n)|$ for some constant $C>0$ .

So, $\sum_{k=1}^n \left( \sqrt{1+k/n^2} -1 \right) = \sum_{k=1}^n \left( \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right) \right) = \frac{n(n+1)}{2} \frac{1}{2n^2} + O\left(n\left( \frac{k}{n^2} \right)^2 \right) = \frac{n(n+1)}{2} \frac{1}{2n^2} + O(1/n)$ where the first equalityfollows from maclaurin series, second equality follows from evaluating the sum of the first term, and the sum containing $n$ terms, and the third equality follows from $k \leq n$.

Now, take the limit as $n \to \infty$ to get $\frac{1}{4}$.