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$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n}$$

Could you please give me a hint how to find this limit?

Martin Sleziak
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Mirak
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2 Answers2

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$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n} =\lim_{n\rightarrow\infty}\dfrac1n\sum_{k=1}^{n}\frac1{\dfrac kn+1}$$

Use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

lab bhattacharjee
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  • $f(x) = \frac{1}{1+x} \Rightarrow \int f(x) = \ln (1+x) = \ln (1+1) - \ln (1) = \ln 2 $. Thanks! Could you tell me how do you figure out that extracting 1/n will lead to integral? – Mirak May 13 '15 at 05:45
  • @Mirak, Observe that extracting $1/n$ gives us $f(k/n)$ – lab bhattacharjee May 13 '15 at 05:47
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$$\sum_{k=1}^n\frac1{k+n}~=~\sum_{k=n+1}^{2n}\frac1k~=~\sum_{k=1}^{2n}\frac1k~-~\sum_{k=1}^n\frac1k~=~H_{2n}-H_n~\simeq~\ln2n-\ln n~=~\ln\frac{2n}n$$ Can you take it from here, and evaluate what happens as $n\to\infty$ ? :-$)$

Lucian
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    I think it's better to use $o(1)$ in this case instead of $\simeq$. $$H_n=\ln(n)+\gamma +o(1)$$ It's generally a bad idea to use $\simeq$ when subtracting two equivalent functions. – Kitegi May 13 '15 at 07:34