# Convergence of Infinite Series

#### Prism

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?

#### Laurent

MHF Hall of Honor
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.

#### Prism

Thank you very much!

#### p0oint

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
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.