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$$\lim_{n \rightarrow \infty} \sum^n_{k=1} \frac {1} {k+n} = \ln 2$$

I'm supposed to find the limit and get the right part as an answer. I don't know what to do with it. Could someone explain?

ilbra
  • 31

3 Answers3

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Hint: Think of Riemann sums and note that $$ \sum\limits_{k=1}^n\frac{1}{k+n}=\sum\limits_{k=1}^n\frac{1}{1+\frac{k}{n}}\frac{1}{n} $$

Norbert
  • 56,803
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Write it as $$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\frac{k}{n}+1}$$

and look it as Riemann sums for the integral $$\int_{0}^{1}\frac{1}{x+1}dx$$

OR.
  • 5,941
3

Let me give another method without using the Riemann sum.

We use the asymptotic expansion $$\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+o(1)$$ where $\gamma$ is the Euler's constant hence $$\sum_{k=1}^n\frac{1}{k+n}=\sum_{k=n+1}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=n}^{2n}\frac{1}{k}=\log(2n)-\log(n)+o(1)=\log 2+o(1)$$ and we deduce the desired result.