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Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?

I need to Prove the following sum converges:

$\lim_{n\to\infty}\sum\limits_{i=1}^n\dfrac{1}{n+i}$

What methods can I use?

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    Language note: This is not a "sum convergence" problem. We say a sum converges of the sequence $a_0$, $a_0+a_1$, $a_0+a_1+a_2$,... converges. This problem is not a problem of that type. Rather this is a question about the limit of a sequence of different (related) sums.2012-09-29

4 Answers 4

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$\sum_{i=1}^n\frac{1}{n+i}=\frac{1}{n}\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\xrightarrow [n\to\infty]{} \int_0^1\frac{dx}{1+x}=...$

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    Riemann sums, as simple as that. Since the function $\,f(x)=\frac{1}{1+x}\,$ is continuous in $\,[0,1]\,$ its Riemann integral exists there and we can thus choose *any* subdivision of $\,[0,1]\,$ we want to form the Riemann sums. We choose $\,\{x_0=0\,,\,x_1=1/n\,,\,x_2=2/n\,,...,\,x_n=n/n=1\}\,$ and we form the Riemann sum $\sum_{i=1}^nf(x_i)(x_i-x_{i-1})=\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\cdot\frac{1}{n}$2012-09-29
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Hint: you can rewrite your sum as $ \lim_{n\to\infty} \frac{1-0}{n}\sum_{i=1}^n \frac{1}{1+\frac{i}{n}}, $ Now do you know the definition of Riemann's integral?

Added: Somehow the english Wiki page doesn't not seem to show this as explicitly as the french one does, but you can have a look at this page, the first section shows what you have, with $f$ replaced by $\frac{1}{1+x}$

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    It is somewhat the way Riemann's integral is constructed in the first calculus course, via approximatio of areas under curves by rectangle.2012-09-29
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HINT: The sequence of sums is bounded above by $1$: $\sum_{i=1}^n\dfrac{1}{n+i}<\sum_{i=1}^n\frac1n=1\;.$

It’s also strictly increasing, as you can show by calculating

$\sum_{i=1}^{n+1}\dfrac{1}{n+1+i}-\sum_{i=1}^n\dfrac{1}{n+i}=\frac1{2(n+1)}+\sum_{i=1}^n\left(\frac1{n+1+i}-\frac1{n+i}\right)\;;$

I’ll leave the rest of that calculation to you. Note that the last sum telescopes.

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    @CodeKingPlusPlus: Exactly.2012-09-29
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Hint

$\frac{1}{n+i}=\frac{1}{n\left(1+\frac{i}{n}\right)}$, then rewrite sum $\sum\limits_{i=1}^n\dfrac{1}{n+i}$ as Riemann sum for appropriate integral.