# Thread: Covergence of series

1. ## Covergence of series

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

2. Can I use $\displaystyle ln(1+\frac{1}{k}) \geq ln(\frac{1}{k})$

And $\displaystyle \displaystyle\sum_{k=1}^{\infty} \; ln(\frac{1}{k})$ is divergent , so my original series is also diverging?

3. $\displaystyle \ln\left(\frac{1}{k}\right)\leq \ln\left(1+\frac{1}{k}\right)$

Study $\displaystyle \ln\left(\frac{1}{k}\right)=\ln(1)-\ln(k)$ and see that it diverges.

EDIT: Uhm, you got it.

4. Hello,
Originally Posted by Twig
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
$\displaystyle 1+\frac 1k=\frac{k+1}{k}$

So $\displaystyle \ln\left(1+\frac 1k\right)=\ln(k+1)-\ln(k)$

So this is a telescoping series and finally, $\displaystyle \sum_{k=1}^\infty \ln\left(1+\frac 1k\right)=\lim_{n\to\infty} \sum_{k=1}^n \ln\left(1+\frac 1k\right)=\lim_{n\to\infty} \ln(n+1)-\ln(1)=\infty$

5. In general the series...

$\displaystyle \sum_{n=1}^{\infty} \ln (1+a_{n})$

... converges if and only if converges the series...

$\displaystyle \sum_{n=1}^{\infty} a_{n}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Hello chisigma,

I thought$\displaystyle \frac {1}{k}$ converges?

7. Originally Posted by calc101
Hello chisigma,

I thought$\displaystyle \frac {1}{k}$ converges?

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.

8. No, I meant the sequence $\displaystyle \frac {1}{k}$ converges to 0?

Not the function $\displaystyle \frac {1}{x}$

9. Yes, the sequence $\displaystyle (a_{n})_{n=1}^{\infty} = \frac{1}{n}$ converges to 0.