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We have

$\lim_{n \to \infty} \frac{1}{n}\Sigma_{k=1}^{n}f(\frac{k}{n})$

and

$f:[0,1]\rightarrow \mathbb{R}$ is continuous function.

How to rewrite this limit into definite integral and than to calculate

$\lim_{n \to \infty} \Sigma_{k=n+1}^{2n}(\frac{1}{k})$ ?

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    the first sum is the defintion of a riemann integral, right? – 2017-01-10
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    For the second question set $d=k-n$ and see what happens as an intermediate step you will find sth like $\int_0^1\frac{1}{1+x}dx$ – 2017-01-10

1 Answers 1

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The first sum is the definition of Riemann Integral in limit form. Therefore$\lim_\limits{n\to\infty}\frac1{n}\sum_{k=1}^{\infty}f(\frac{k}{n})=\int_0^1f(x)dx$. Now, the required sum$=\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+k}=\lim_{n\to\infty}\frac1{n}\sum_{k=1}^n\frac1{1+(\frac{k}{n})}=\int_0^1\frac1{1+x}dx=\ln 2$