1. ## Telescoping Series

Please write out how you did this as well. I'd like to understand and not just know the answer. Thanks

$\displaystyle \sum\limits_{n = 1}^\infty {ln (1+ (1/n))}$

2. Originally Posted by geniusboy91
Please write out how you did this as well. I'd like to understand and not just know the answer. Thanks

Σ from n=1 to ∞ of ln(1+(1/n))
Just make a common denominator

$\displaystyle \sum_{n \geq 1} \left[ ln(1+\frac{1}{n})=ln(\frac{n+1}{n})=ln(n+1)-ln(n) \right]$

I think you can take it from here.

3. Ahh!!! I knew it would be some simple algebraic thing I was missing. And so that would lead to the sequence of patial sums being

ln(n+1) - ln(1) = ln(n+1)

and diverges because:

$\displaystyle \lim_{n\to\infty} {ln(n+1)}$ is infinity?

4. Originally Posted by geniusboy91
Ahh!!! I knew it would be some simple algebraic thing I was missing. And so that would lead to the sequence of patial sums being

ln(n+1) - ln(1) = ln(n+1)

and diverges because:

$\displaystyle \lim_{n\to\infty} {ln(n+1)}$ is infinity?
Correct.
Since the sequence of the partial sums diverges then ,by the definition, the series diverges.