Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have fo find: $$\displaystyle\lim_{n\to\infty}\frac{1}{n}\Bigg(1+\frac{2}{1+\sqrt{2}}+\frac{3}{1+\sqrt{2}+\sqrt{3}}+\cdots+\frac{n}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}\Bigg)$$

using the Cesaro-Stolz Theorem.

Applying what the theorem states once i get: $\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}$

My question is: can i apply it again to get: $\displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n+1}}=0$?

share|improve this question
    
Yes, why not? If this new exercise was a completely new exercise, wouldn't you had solved it with S-C? Then, why would the fact that this is not a new exercise, but part of a larger one change your technique? –  N. S. Jan 4 '13 at 21:33
    
Well in school they never tell us what freedom we have while doing math, so i was unsure. I thought logically its possible, i just wanted to be sure. –  phi Jan 4 '13 at 21:36
1  
BTW if you use SC, the new limit should be $\lim_n \frac{1}{\sqrt{n+1}}$. –  N. S. Jan 4 '13 at 21:38

1 Answer 1

up vote 1 down vote accepted

S-C is the easiest way to calculate

$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}$$

If you want an alternate solution, note that

$$1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}=\sqrt{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} \,,$$

and

$$\lim_n \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} =\int_0^1 \sqrt{x} dx \,.$$

Then

$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}=\displaystyle\lim_{n\to\infty} \frac{n+1}{n\sqrt{n}}\frac{1}{ \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} }=0$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.