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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.

Oran Sherf
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2 Answers2

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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

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

user
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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: $$\frac{\sqrt{2n+1}-1}{2\sqrt{n}}$$

King Tut
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  • I didn't got it how you got to the last sum, and how exact the sigma "gone" – Oran Sherf Apr 06 '18 at 13:04
  • @oran its the magic of telescope. Write a few terms. Numerator will have $\sqrt{3}-1 + (\sqrt 5 - \sqrt{3}) + (\sqrt{7}- \sqrt 5)$ ... In each step, intermediate terms get cancelled and only end terms remain – King Tut Apr 06 '18 at 13:10