Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$

Question

I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.

I tried to use the squeeze theorem, failed.
I tried to use a sequence defined recursively: $a_{n+1} = {a_n} + \frac{1}{\sqrt{(n+1)^2 +n+1}}$. It is a monotone growing sequence, for every $n$, $a_n > 0$. I also defined $f(x) = \frac{1}{\sqrt{(x+1)^2 +x+1}}$. So $a_{n+1} = a_n + f(a_n)$. But I'm stuck.

How can I calculate it?

Can you elaborate on your squeezing? It should work. – Git Gud Feb 26 '17 at 18:26 — Git Gud, Feb 26 '17 at 18:26

score 4 · Accepted Answer · answered Feb 26 '17 at 18:32

4

It looks squeezable.

\begin{align} \frac{n}{\sqrt{n^2+n}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le \frac{n}{\sqrt{n^2+1}} \\ \\ \frac{1}{\sqrt{1+\frac{1}{n}}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le \frac{1}{\sqrt{1+\frac{1}{n^2}}} \end{align}

answered Feb 26 '17 at 18:32

zahbaz

10,441

score 1 · Answer 2 · answered Feb 26 '17 at 18:29

If $k\in[1,n]$ then the difference between $\frac{1}{\sqrt{n^2+k}}$ and $\frac{1}{n}$ is rather small: $$ 0\leq \frac{1}{n}-\frac{1}{\sqrt{n^2+k}} = \frac{k}{n\sqrt{n^2+k}(n+\sqrt{n^2+k})}\leq \frac{1}{2n^2} $$ hence $\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$ tends to $1$ as $n\to +\infty$, since $\sum_{k=1}^{n}\frac{1}{n}=1$ and $0\leq\sum_{k=1}^{n}\frac{1}{2n^2}=\frac{1}{2n}$.

score 1 · Answer 3 · answered Feb 26 '17 at 18:35

1

Hint: $$n\frac{1}{\sqrt{n^2+n}}\leq\sum\frac{1}{\sqrt{n^2+n}}\leq n\frac{1}{\sqrt{n^2}}$$

answered Feb 26 '17 at 18:35

Nosrati

29,995

Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$

3 Answers3