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I have a homework question that is as follows:

Calculate

$$\lim _{ n\to \infty } \left[ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \right] $$

by writing the expression in brackets as

$$\frac { 1 }{ n } \left[ \frac { 1 }{ 1+1/n } +\frac { 1 }{ 1+2/n } +...+\frac { 1 }{ 1+n/n } \right] $$

and recognizing the latter as a Riemann sum.

I am aware of what a Riemann sum is, but not quite sure what the first expression is depicting the sum of. The second expression makes almost no sense to me and I am not sure what the question is general is trying to get me to do. Any help would be greatly appreciated as I do not directly want the answer, just examples and guiding steps towards being able to solve it myself. Thanks!

haqnatural
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ThoseKind
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    Are you sure you want limit as $x$ approaches infinity? not the limit as $n$ approaches infinity? $x$ has nothing to do with what is inside of it, so it does not change it at all. – JMoravitz Aug 27 '16 at 21:39
  • Perhaps it was asking for you to take the limit as $n\to\infty$ instead of $x$? And all it did for the step was factor out a $1/n$ term from everything. Then evaluating the limit should be easier. – Simply Beautiful Art Aug 27 '16 at 21:40
  • The exercise is an example in how to rewrite an expression to the form of a Riemann sum and allow you to evaluate its limit as an integral. The goal is train you to detect this sort of pattern in other sums. Once you become familiar with this, other similar limit of sums, say $\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k^2}$ will become a piece of cake. – achille hui Aug 27 '16 at 22:23
  • Have a look at: http://math.stackexchange.com/q/498040/321264, http://math.stackexchange.com/q/1396839/321264 – StubbornAtom Aug 28 '16 at 08:03

3 Answers3

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$$\lim _{ n\to \infty } \left[ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \right] =\lim _{ n\to \infty } \frac { 1 }{ n } \left[ \frac { 1 }{ 1+\frac { 1 }{ n } } +\frac { 1 }{ 1+\frac { 2 }{ n } } +...+\frac { 1 }{ 1+\frac { n }{ n } } \right] =\\ =\lim _{ n\to \infty } \frac { 1 }{ n } \sum _{ k=1 }^{ n }{ \frac { 1 }{ 1+\frac { k }{ n } } } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+x } dx } =\color{red}{\ln 2} $$

haqnatural
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  • Would you be able to provide some background as to how you got this far? I understand the summation notation but am not sure as to how you got to the bounds on the integral in the next step, or how you got the answer of ln 2. – ThoseKind Aug 27 '16 at 22:53
  • We know that $\lim { n\rightarrow \infty }{ \frac { b-a }{ n } \sum _{ k=1 }^{ n }{ f\left( a+k\frac { b-a }{ n } \right) } } =\int _{ a }^{ b }{ f\left( x \right) dx } $ in our case $f:\left[ 0,1 \right] \rightarrow \mathbb{R}$ which is divided to n pieces and note that $\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+x } ={ \int _{ 0 }^{ 1 }{ \frac { d\left( x+1 \right) }{ 1+x } = } \left[ \ln { \left| 1+x \right| } \right] }{ 0 }^{ 1 } } =\ln { \left( 1+1 \right) -\ln { \left( 1+0 \right) =\ln { 2 } } } $ – haqnatural Aug 27 '16 at 23:13
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Hint:

Write the second expression as $$\frac1n\sum_{k=1}^n\frac1{1+\frac kn}=\frac1n\sum_{k=1}^n\frac1{1+x_k}$$ if we set $x_k=\frac kn$.

Bernard
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Beside the Riemann sum, you also approach the problem using harmonic numbers $$S_n=\sum_{i=1}^n\frac 1{n+i}=H_{2 n}-H_n$$ and using the asymptotics $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ you will get $$S_n=\sum_{i=1}^n\frac 1{n+i}=H_{2 n}-H_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit as how it is approached.

This would give good approximate values for finite values of $n$. For example $$S_{10}= \frac{155685007}{232792560}\approx 0.6687714$$ while $$\log(2)-\frac{1}{40}+\frac{1}{1600}\approx 0.6687722$$

Claude Leibovici
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