# Convergence of Infinite Series

• Jun 7th 2010, 01:06 PM
Prism
Convergence of Infinite Series
Hi there,

I'm trying to prove that if

An <= (1 - 1/n^a)^n

(when 0 < a < 1)

Then the series ∑An converges.

I managed to get to the form of

∑ (1/e)^(n^b)

(where 0<b<1)

But I can't even prove that this more simple series converges.

• Jun 7th 2010, 02:33 PM
Laurent
Quote:

Originally Posted by Prism
I managed to get to the form of

∑ (1/e)^(n^b)

(where 0<b<1)

But I can't even prove that this more simple series converges.

For instance, you can say that $\frac{n^2}{e^{n^b}}=\frac{(n^b)^{2/b}}{e^{n^b}}\to 0$ because $\frac{x^{2/b}}{e^x}\to_{x\to\infty} 0$. Therefore, for large $n$, we have $\frac{n^2}{e^{n^b}}\leq 1$, thus $0<\frac{1}{e^{n^b}}\leq \frac{1}{n^2}$ hence the convergence.
• Jun 8th 2010, 10:58 AM
Prism
Thank you very much!
• Jun 8th 2010, 01:35 PM
p0oint
I assume the original problem is:
If $A_n \leq (1 - \frac{1}{n^a})^n$ where 0 < a < 1 then ∑An converges.

Just another approach:

Prove that $\sum_{n=1}^{\infty} (1 - \frac{1}{n^a})^n$ is convergent then by the comparison test also ∑An converges.

or it is the same approach?
• Jun 8th 2010, 04:09 PM
Drexel28
Quote:

Originally Posted by p0oint
I assume the original problem is:
If $A_n \leq (1 - \frac{1}{n^a})^n$ where 0 < a < 1 then ∑An converges.

Just another approach:

Prove that $\sum_{n=1}^{\infty} (1 - \frac{1}{n^a})^n$ is convergent then by the comparison test also ∑An converges.

or it is the same approach?

I would note that $\left(1-\frac{1}{n^\alpha}\right)^n=e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}$. Then, for sufficiently large $n$ we have that $n^\alpha>1\implies \frac{1}{n^\alpha}<1$ and thus $\ln\left(1-\frac{1}{n^\alpha}\right)=-\sum_{m=1}^{\infty}\frac{1}{n^{\alpha m}m}$ which clearly behaves as $\frac{-1}{n^\alpha}$ for large $n$ and thus $\left(1-\frac{1}{n^\alpha}\right)\overset{n\to\infty}{\sim }e^{-n^{1-\alpha}}$....which I just realized is what you did and Laurent finished. My apologies.