Hi
I want to determine if this series is convergent or divergent.
Can anyone give me a hint on which theorem to use here?
$\displaystyle \displaystyle\sum_{k=1}^{\infty} \; ln(1+\frac{1}{k}) $
thx
Hello,
$\displaystyle 1+\frac 1k=\frac{k+1}{k}$
So $\displaystyle \ln\left(1+\frac 1k\right)=\ln(k+1)-\ln(k)$
So this is a telescoping series and finally, $\displaystyle \sum_{k=1}^\infty \ln\left(1+\frac 1k\right)=\lim_{n\to\infty} \sum_{k=1}^n \ln\left(1+\frac 1k\right)=\lim_{n\to\infty} \ln(n+1)-\ln(1)=\infty$
Itīs diverging. The n:th term approaches zero as n approaches infinity, but
$\displaystyle \displaystyle\sum_{k=1}^{\infty} \frac{1}{k} $ is diverging.
You know about the integral test?
$\displaystyle \displaystyle\lim_{R\to \infty} \int_{1}^{R} \frac{1}{x} \, dx = ln(R)-ln(1) = \infty $
Note: $\displaystyle \displaystyle\sum_{k=1}^{\infty} \frac{1}{k^{1.000001}} $ is convergent.