I have question which to show that sum(1/n)log[1+(1/n)] is convergent. I don't think i can use integral or ratio or raabe's test. I am guessing I have to use limit comparison test ??? Help me !

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- May 13th 2011, 07:28 AMmathsohardconvergence test
I have question which to show that sum(1/n)log[1+(1/n)] is convergent. I don't think i can use integral or ratio or raabe's test. I am guessing I have to use limit comparison test ??? Help me !

- May 13th 2011, 07:33 AMTheEmptySet
- May 13th 2011, 07:40 AMmathsohard
I haven't learned l'Hospitals rule yet lol is that the only way ???

- May 13th 2011, 07:48 AMTheEmptySet
Well if you do the limit comparison you will need to calculate the limit

$\displaystyle \frac{\frac{\log\left( 1+\frac{1}{n}\right)}{n}}{\frac{1}{n^2}}=n \log \left( 1+\frac{1}{n} \right) $

This is an indeterminate form of infinity times zero. What methods have you learned to resolve these types of limits? The only other method I can think of is to use a Taylor series expansion, but that is usually covered after the topics you are learning right now. - May 13th 2011, 08:29 AMmathsohard
Well can you show me how to do with l'Hospitals rule then???

- May 13th 2011, 08:50 AMTheEmptySet
Here is a link to the theorem:

L'Hôpital's rule - Wikipedia, the free encyclopedia

$\displaystyle \frac{\frac{\log\left( 1+\frac{1}{n}\right)}{n}}{\frac{1}{n^2}}=\frac{ \log \left( 1+\frac{1}{n} \right)}{\frac{1}{n}} $

This is of the form zero over zero so we take the derivative of the numerator and denominator to get

$\displaystyle \lim_{n \to \infty}\frac{\frac{1}{ 1+\frac{1}{n}}\cdot \frac{-1}{n^2}}{\frac{-1}{n^2}}=1$