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I really have difficulties with Riemann Sums, especially the ones as below:

$$\lim_{n\to\infty} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n}\right)$$ When i try to write this as a sum, it becomes $$\frac { 1 }{ n } \sum _{ k=1 }^{ 2n } \frac { 1 }{ 1+\frac { k }{ n } } .$$ The problem is, however, to be able to compute this limit as an integral I need to have this sum from $1$ to $n$. There are some other questions like this, but if I can understand it, I will be able solve others.

Michael Hardy
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Hckr
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    Hint: Do some multiplications by $\frac{2}{2}$, then do a replacement $m=2n$. – MartianInvader Jan 02 '14 at 22:03
  • I couldn't get it. Could you share a solution to the limit above as an answer, please? – Hckr Jan 02 '14 at 22:06
  • Now set $x_k=\frac{k}{n}$ and $dx=\frac1n$ and look again. – Lutz Lehmann Jan 02 '14 at 22:10
  • If you look at the question, it cannot be applied here yet, because the sum is from 1 to 2n not n. – Hckr Jan 02 '14 at 22:12
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    What makes this a little unusual is that your sum has 2n terms instead of n terms, so you're dividing the interval $[0,2]$ into m equal intervals and then letting $m=2n$, as suggested in the hint above. – user84413 Jan 02 '14 at 22:29

3 Answers3

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$$ \frac{1}{n} \sum_{k=1}^{2n} \frac{1}{1+\frac{k}{n}}= \sum f(x_k)\,\Delta x. $$ where $f(x) = \dfrac{1}{1+x}$ and $\Delta x = \dfrac 1 n$. The variable goes from $1/n\to0$ to $(2n)/n=2$. Hence the sum approaches $$ \int_0^2 \frac{1}{1+x}\,dx. $$

Michael Hardy
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  • if the integral is from 0 to 2, then Δx should be equal to (2-0)/n = 2/n? – Hckr Jan 02 '14 at 22:17
  • If you divide the interval from $0$ to $2$ into $2n$ non-overlapping subintervals of equal lengths, then each subinterval has length $1/n$. – Michael Hardy Jan 02 '14 at 22:19
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With Eulero-Mascheroni : $$\sum_{k = 1}^{n}\frac{1}{k} - \log{n} \rightarrow \gamma$$ $$\sum_{k = 1}^{3n}\frac{1}{k} - \log{3n} \rightarrow \gamma$$ so $$\sum_{k = n+1}^{3n}\frac{1}{k} - \log{3n} +\log{n} \rightarrow 0$$ and then$$\sum_{k = n+1}^{3n}\frac{1}{k} \rightarrow \log{3}$$

WLOG
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We can rewrite our expression as $$\frac{1}{2n}\sum_{k=1}^{2n}\frac{2n}{n+k}$$ and then as $$\frac{1}{2n}\sum_{k=1}^{2n}\frac{1}{\frac{1}{2}+\frac{k}{2n}}.$$ This is a Riemann sum for $$\int_0^1 \frac{1}{\frac{1}{2}+x}\,dx.$$ The integral, and therefore the desired limit, is $\ln(3/2)-\ln(1/2)$, or more simply $\ln 3$.

André Nicolas
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