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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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marked as duplicate by Marvis, Thomas, Martin Sleziak, J. M., Chris Eagle Oct 5 '12 at 16:01

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Comparison of sums and integrals. –  Did Sep 29 '12 at 20:59
    
Are you taking the limit for $n\to \infty$? –  Jean-Sébastien Sep 29 '12 at 21:00
    
Yes, I will specify that –  CodeKingPlusPlus Sep 29 '12 at 21:01
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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. –  Thomas Andrews Sep 29 '12 at 21:24
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4 Answers 4

up vote 3 down vote accepted

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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What does this last equation show? You are finding the $(n+1)th$ term correct? –  CodeKingPlusPlus Sep 29 '12 at 21:59
    
@CodeKingPlusPlus: No, if $s_n$ is the $n$-th term, I’m calculating $s_{n+1}-s_n$. What kind of result would show that the sequence of $s_n$’s is increasing? –  Brian M. Scott Sep 29 '12 at 22:01
    
Ok, so if the difference of $s_{n+1} - s_n > 0$ then $s_n$ is increasing. –  CodeKingPlusPlus Sep 29 '12 at 22:07
    
@CodeKingPlusPlus: Exactly. –  Brian M. Scott Sep 29 '12 at 22:09
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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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Is Riemann's integral that theorem they teach you in calc 1 when you take the number of rectangles under a curve to be infinite and compute their areas? If so, it has been a long long time! hahah –  CodeKingPlusPlus Sep 29 '12 at 21:09
    
It is somewhat the way Riemann's integral is constructed in the first calculus course, via approximatio of areas under curves by rectangle. –  Jean-Sébastien Sep 29 '12 at 21:12
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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.

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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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Could you show some derivation of how you get $\int\dfrac{dx}{1+x}$ –  CodeKingPlusPlus Sep 29 '12 at 21:34
    
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 $$\sum_{i=1}^nf(x_i)(x_i-x_{i-1})=\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\cdot\frac{‌​1}{n}$$ –  DonAntonio Sep 29 '12 at 21:41
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