Calculate a limit using sandwich rule for sequences

Question

I'm trying to find the limit of sequence by using sandwich rule for sequences:

$d = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}$

After a couple of operations i got to:

$n[\frac{1}{\sqrt{n}}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}] ≤ \frac{1}{\sqrt{n}}\frac{1}{\sqrt{3}+\sqrt{1}} + \frac{1}{\sqrt{n}}\frac{1}{\sqrt{5}+\sqrt{3}} + ... + \frac{1}{\sqrt{2n+1}+\sqrt{2n-1}}≤ n[\frac{1}{\sqrt{n}}\frac{1}{\sqrt{3}+\sqrt{1}}]$

For calculate the limit by the sandwich theorem, I've calculated the limits for both of the sides and it were different.

I'll be glad to know where is my mistake here.

Best regards.

Answer 1

Note that yours is a telescoping sum $$d = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}} = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{\sqrt{2k+1}-\sqrt{2k-1}}{2}$$

The intermediate terms cancel and only last and first terms are left. The sum therefore is: $$\frac{\sqrt{2n+1}-1}{2\sqrt{n}}$$

Answer 2

HINT

As an alternative by Stolz-Cesaro

$$\frac{a_n}{b_n}=\frac{\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}}{\sqrt{n}}$$

we have

$$\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{2n+3}+\sqrt{2n+1}}}{\sqrt{n+1}-\sqrt{n}}$$