# Convergence of a series

#### Gok2

Hi everyone.
I need to prove that if 0 < a < 1 , then the series $$\displaystyle \sum_{n=1}^{\infty} (1-\frac{1}{n^a})^n$$ converges.
I tried to use the limit test and the comparison test but I couldn't manage to.
Can someone give me a hint how to prove that?

#### Drexel28

MHF Hall of Honor
Hi everyone.
I need to prove that if 0 < a < 1 , then the series $$\displaystyle \sum_{n=1}^{\infty} (1-\frac{1}{n^a})^n$$ converges.
I tried to use the limit test and the comparison test but I couldn't manage to.
Can someone give me a hint how to prove that?

Notice that $$\displaystyle \left(1-\frac{1}{n^\alpha}\right)^n=e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}$$. But, notice that since $$\displaystyle 0<\frac{1}{n}\leqslant 1$$ we may apply the series to the natural log to obtain that $$\displaystyle e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}=e^{-n\left(\frac{1}{n^\alpha}+\frac{1}{2n^{2\alpha}}+\cdots\right)}\approx e^{-n^{1-\alpha}}$$ so what happens if $$\displaystyle 0<\alpha<1$$?