Could you explain how to find this limit?
$\lim_{n \rightarrow \infty} \sum^n_{k=1} \frac{1}{k+n}$
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1You could try to rewrite the sum to look like a Riemann sum. – Harald Hanche-Olsen Jul 13 '14 at 14:30
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What do you mean? Riemann sum is approximation for integral – EvanTello Jul 13 '14 at 14:37
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@Thisismuchhealthier. Could you explain the transformation that makes these two problems equivalent? For now, I can't see how they're duplicates. – apnorton Jul 13 '14 at 16:16
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1@anorton In the question here, let $m=k+n$. The sum becomes $\sum_{m=n+1}^{2n} \frac{1}{m}$. – Jul 13 '14 at 16:18
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@Thisismuchhealthier. AH! I can't believe I didn't see that. Nice... :) (And voted.) – apnorton Jul 13 '14 at 16:20
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$\lim_{n\rightarrow \infty } \sum_{k=1}^n \frac{1}{k+n}=\lim_{n\rightarrow \infty } \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}\cdot \frac{1}{n}=\int_0^1 \frac{dx}{x+1}=\ln 2$.
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1@EvanTello think in the reverse direction. What will the Riemann sum associated with the integral in RHS looks like? In this particular case, the LHS actually equal to the lower and the right Riemann sum associated with the RHS for the partition of $[0,1]$ into $n$ sub-intervals. – achille hui Jul 13 '14 at 14:52
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