Convergence of a series

Jan 2010
21
0
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?

Thanks in advance!
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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?

Thanks in advance!
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 \(\displaystyle 0<\alpha<1\)?