# Limit of sum of exponential

1. Jan 6, 2008

### dakongyi

1. The problem statement, all variables and given/known data
consider {sum from k=0 to n of e^(sqrt(k))}/{2sqrt(n)e^(sqrt)}.how to prove that the limit when n approaches infinity is 1?
or in latex form,
\lim_{n \to \infty}\frac{\sum_{k=0}^{n}e^{\sqrt{k}}}{2\sqrt{n}e^{\sqrt{n}}}=1

2. Relevant equations
Nil

3. The attempt at a solution
I tried to use logarithm to remove the exponential, but failed.

Last edited: Jan 7, 2008
2. Jan 7, 2008

### Gib Z

Welcome to Physicsforums!

That expression unfortunately is very hard to read in that form :( To get the LaTeX working on this forum, you must use the [ tex ] and [ /tex ] tags, without the spaces.

I'll do that now, then try to help =]

$$\lim_{n \to \infty}\frac{\sum_{v=0}^{n}e^{\sqrt{k}}}{2\sqrt{n} e{\sqrt{n}}}=1$$

EDIT: Ok it seems you meant;

$$\lim_{n \to \infty}\frac{\sum_{k=0}^{n}e^{\sqrt{k}}}{2\sqrt{n} e^{\sqrt{n}}}=1$$

3. Jan 7, 2008

### dakongyi

thanks for the help. i need the second equation, thanks for the edit :)

4. Jan 7, 2008

### dakongyi

it seems that the $$2\sqrt{n}$$ comes from the derivative of $$e^\sqrt{n}$$, but i couldn't think of anyway to make use of differentiation.

5. Jan 7, 2008

### Dick

Replace $$\sum_{k=0}^{n}e^{\sqrt{k}}$$ with $$\int_{0}^{n} e^{\sqrt{x}}dx$$ since it has the same behavior for large n (just like in the integral test for convergence). Now use l'Hopital.

6. Jan 7, 2008

### Defennder

Could you explain how this step is justified?

7. Jan 7, 2008

### CompuChip

Actually it seems to me that
$$\sum_{k=0}^{n}e^{\sqrt{k}} - \int_{0}^{n} e^{\sqrt{x}}dx$$
is diverging

8. Jan 7, 2008

### Dick

Actually, I was hoping no one would ask for a detailed justification. Yes, the difference probably is divergent. But I think the denominator is even more divergent. One would have to show that the difference is small compared to the denominator. I'll try and come up with a good argument in a bit.

9. Jan 7, 2008

### Dick

Ok, try this. Apply l'Hopital first, using finite differences instead of derivatives. The difference between the nth partial sum and the (n-1)th partial sum in the numerator is e^(sqrt(n)). The difference between the value of the denominator at n and at n-1 is 2sqrt(n)e^sqrt(n)-2sqrt(n-1)e^sqrt(n-1). For large n I would approximate the denominator using a difference quotient for the derivative of 2sqrt(n)e^sqrt(n). There. That's the same thing, except I'm not making any claim that the integral 'approximates' the sum.

10. Jan 7, 2008

### Dick

Ok, now you're going ask how do I know the difference quotient can be approximated by a derivative. Skip that. Just directly show that the differences above approach the desired limit. It's the same machinery you'd use to show the difference quotient can be approximated by a derivative.

11. Jan 7, 2008

### dakongyi

guess you are using Stolz-Cesàro theorem. oh, now i need to prove that theorem? :rofl: thanks for the help, i will try to prove that Stolz-Cesàro theorem...

12. Jan 7, 2008

### dakongyi

if i apply what you said, i will need to prove
$$\lim_{n \to \infty}\frac{e^{\sqrt{n}}}{2\sqrt{n}e^{\sqrt{n}}-2\sqrt{n-1}e^{{\sqrt{n-1}}}}=1$$
which gives
$$\lim_{n \to \infty}\frac{1}{2\sqrt{n}-2\sqrt{n-1}e^{{\sqrt{n-1}-\sqrt{n}}}}=1$$
it then suffies to show that
$$\lim_{n \to \infty}{2\sqrt{n}-2\sqrt{n-1}e^{{\sqrt{n-1}-\sqrt{n}}}}=1$$
i couldn't proceed from here. any hints?

Last edited: Jan 7, 2008
13. Jan 7, 2008

### Dick

sqrt(n-1)-sqrt(n)=-1/(sqrt(n-1)+sqrt(n)). Expand the exponential to first order. Oh, yeah, Stolz-Cesaro, that one. I figured it must have a name, but I didn't know it.

14. Jan 7, 2008

### dakongyi

I got it. A million thanks to those who helped.

15. Jan 8, 2008