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Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?

Please help me with this homework question.

Let $\lbrace a_{n} \rbrace$ be the sequence defined by $$a_{n} = \frac{1}{n+1} + \frac{1}{n+2} ... +\frac{1}{2n}$$ for each positive integer $n$. Prove that this sequence converges to ln 2 by showing that $a_{n}$ is related to the partial sums of the series $\displaystyle\sum\limits_{k=0}^\infty (-1)^{k+1}/k.$

I know that $\displaystyle\sum\limits_{k=0}^\infty (-1)^{k+1}/k$ converges to ln 2.

I also know that

$s_{1} = 1,$

$s_{2} = 1/2,$

$s_{3} = 5/6,$

$s_{4} = 7/12,$

etc.

However, I am having a hard time understanding how $\lbrace a_{n} \rbrace$ is related to the partial sum of the given series. Any help or hints are greatly appreciated.

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    You can find multiple (other) proofs in the answers here: http://math.stackexchange.com/questions/113401/limit-of-u-n-sin-frac1n1-cdots-sin-frac12n2012-02-29

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