What is the Limit of Sum of Exponential as n Approaches Infinity?

[dakongyi]

Homework Statement

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

Homework Equations

Nil

The Attempt at a Solution

I tried to use logarithm to remove the exponential, but failed.

 

[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

 

[dakongyi]

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

 

[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.

 

[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.

 

[Defennder]

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.

Could you explain how this step is justified?

 

[CompuChip]

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

 

[Dick]

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

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.

 

[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.

 

[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.

 

[dakongyi]

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

 

[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?

 

[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.

 

[dakongyi]

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

 

[Gib Z]

I think one can also do this with the Euler-Maclaurin formula.
http://en.wikipedia.org/wiki/Euler-Maclaurin_formula