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$$ \lim_{n\to \infty} \left(\frac{1}{n+1} + \frac{1}{n+2} + ..+ \frac{1}{2n}\right)$$

How do i find the limit by expressing it as a definite integral of an appropriate function via Riemann Sums?

I do know that for riemann sums

$\Delta x = \frac{b-a}{n}$ and $ x_i^* = a + \frac{b-a}{n}$

So i would need something like this

$$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}\cdot f\left(a+i\frac{b-a}{n} \right)$$

But how should i go about solving the question?

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marked as duplicate by Jyrki Lahtonen, Najib Idrissi, user99914, Hakim, Davide Giraudo Nov 2 '14 at 11:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ OMG!! I'm competing with $10K$ users! $\endgroup$ – user171358 Nov 2 '14 at 8:32
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Notice That : $$\frac{1}{n+r}=\frac{1}{n}\frac{1}{1+\frac{r}n}$$

So our some can be written as $$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r} = \lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}$$

Now by replacing $\frac{1}{n}$ with $dx$,$\frac{r}{n}$ with $x$ and $\sum$ with $\int$

$$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}= \lim_{n\to \infty}\frac{1-0}{n}\sum_{r=1}^{n} \frac{1}{1+\left(0+1\cdot\frac{r}{n}\right)}$$ Now comparing this with the form you want we have $b=1$ , $ a=0$ and $f\left(x\right)=\frac{1}{1+x}$

$$\lim_{n\to \infty}\frac{1-0}{n}\sum_{r=1}^{n} \frac{1}{1+\left(0+1\cdot\frac{r}{n}\right)}=\int_0^1f(x)dx=\int_0^1\frac{1}{1+x}dx$$ And resultant integral evaluates to $$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r}=\int_0^1\frac{1}{1+x}dx=\ln\left(\frac{2}{1}\right)=\ln(2)$$

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  • $\begingroup$ how do u get $\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r}=\int_0^1\frac{1}{1+x}dx=\ln\left(\frac{2}{1}\right)=\ln(2)$? $\endgroup$ – Helpisneeded Nov 2 '14 at 8:38
  • $\begingroup$ how do you do that? is there a specific formula or? $\endgroup$ – Helpisneeded Nov 2 '14 at 8:42
  • $\begingroup$ i think i get it. thank you! $\endgroup$ – Helpisneeded Nov 2 '14 at 9:26
  • $\begingroup$ erm what do you mean by accepting answers? There's only one answer no? or am i forgetting something? $\endgroup$ – Helpisneeded Nov 2 '14 at 10:05
  • $\begingroup$ ah i see. thats the answer you're talking about. $\endgroup$ – Helpisneeded Nov 2 '14 at 10:17
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Write the given sum in this way

$$\frac1n\sum_{k=1}^n\frac{1}{1+\frac kn}\xrightarrow{n\to\infty}\int_0^1\frac{dx}{1+x}$$

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Alternatively, $\displaystyle \sum_{k=1}^n \dfrac{1}{k+n} = H_{2n} - \ln (2n) - \left(H_n - \ln n\right) + \ln 2 \to \gamma - \gamma + \ln 2 = \ln 2$

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Hint. Write $$ \frac{1}{n+1} + \frac{1}{n+2} + ..+ \frac{1}{2n}=\frac1n \sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}} $$

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