Convergence of Infinite Series

Jun 2010
3
0
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.

Could someone help me please?

Thanks in advance!
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
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 \(\displaystyle \frac{n^2}{e^{n^b}}=\frac{(n^b)^{2/b}}{e^{n^b}}\to 0\) because \(\displaystyle \frac{x^{2/b}}{e^x}\to_{x\to\infty} 0\). Therefore, for large \(\displaystyle n\), we have \(\displaystyle \frac{n^2}{e^{n^b}}\leq 1\), thus \(\displaystyle 0<\frac{1}{e^{n^b}}\leq \frac{1}{n^2}\) hence the convergence.
 
Nov 2009
130
24
I assume the original problem is:
If \(\displaystyle A_n \leq (1 - \frac{1}{n^a})^n\) where 0 < a < 1 then ∑An converges.

Just another approach:

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

or it is the same approach?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
I assume the original problem is:
If \(\displaystyle A_n \leq (1 - \frac{1}{n^a})^n\) where 0 < a < 1 then ∑An converges.

Just another approach:

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

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