azatkgz
Nov19-07, 10:35 AM
1. The problem statement, all variables and given/known data
Determine whether the series converges absolutely,converges conditionally or diverges.
\sum_{n=1}^{\infty}\ln\left(1+\frac{(-1)^n}{n^p}\right)
where p is a some parameter
3. The attempt at a solution
\ln\left(1+\frac{(-1)^n}{n^p}\right)=\frac{(-1)^n}{n^p}-\frac{1}{n^{2p}}+\frac{(-1)^{3n}}{3n^{3p}}+O(\frac{1}{n^{4p}})
Here
\sum_{n=1}^{\infty}\frac{1}{n^{2p}} converges for p>1/2
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^p} converges absolutely for p>1
My answer is the series converges for p>1/2
for \frac{1}{2}<p\leq 1 it converges conditionally
for p>1 it converges absolutely
Determine whether the series converges absolutely,converges conditionally or diverges.
\sum_{n=1}^{\infty}\ln\left(1+\frac{(-1)^n}{n^p}\right)
where p is a some parameter
3. The attempt at a solution
\ln\left(1+\frac{(-1)^n}{n^p}\right)=\frac{(-1)^n}{n^p}-\frac{1}{n^{2p}}+\frac{(-1)^{3n}}{3n^{3p}}+O(\frac{1}{n^{4p}})
Here
\sum_{n=1}^{\infty}\frac{1}{n^{2p}} converges for p>1/2
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^p} converges absolutely for p>1
My answer is the series converges for p>1/2
for \frac{1}{2}<p\leq 1 it converges conditionally
for p>1 it converges absolutely