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))} $
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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.
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?
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.
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