# Calculating the limit of some sum

I want to calculate the following limit $$\lim_{n\to\infty}\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2n}\right)=?$$ Obviously the result is bounded between $1/2$ and $1$, but how can I calculate the exact result?

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Isn't $\sum_{i=1}^n 1/i$ asymptotic to $\log(n)$ as $n$ gets large? – Stromael Nov 24 '13 at 12:59
I'm not sure how this helps. The term in the limit is $\sum_{i=n}^{2n}1/i$ and not what you wrote. – user91011 Nov 24 '13 at 13:05
1) $|\sum_{i=1}^n1/i-(\gamma+\log(n))|\rightarrow0$ as $n\rightarrow\infty$, where $\gamma$ is the Euler-Mascheroni constant (its value is unimportant, what matters is that it is constant). 2) Can you re-write your sum in terms of harmonic numbers (i.e. partial sums of the form I've given)? – Stromael Nov 24 '13 at 13:09

Rewrite the given sum as follows:

\begin{align}\sum_{k=0}^{n}\frac{1}{n}\frac{1}{1+\frac{k}{n}}\end{align}

Consider the function $f(x)=\frac{1}{1+x}$, for $x\in[0,1]$.

Take the partition $x_0=1$,$x_k=\frac{k}{n}$. In the limit, we find

\begin{align}\int_{0}^{1}f=\lim_{n\rightarrow\infty}\sum_{k=0} ^{n}f(x_k)(x_{k+1}-x_{k})=\lim_{n\rightarrow\infty} \sum_{k=0}^{n}\frac{1}{n}\frac{1}{1+\frac{k}{n}}.\end{align}

So, \begin{align}\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{n+k}=\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{n}\frac{1}{1+\frac{k}{n}}=\int_{0}^{1}f=\int_{0}^{1}\frac{1}{1+x}\text{d}x=\log(2).\end{align}

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Do you mean $\infty$ or $n$ in your first and corresponding sums? – Stromael Nov 24 '13 at 13:12
thank you! I will fix – SnowAngel6147 Nov 24 '13 at 13:15