# Evaluate $\lim_{n\to\infty}\sum_{r=1}^{n}\sin\frac{r\pi}{n}$

Evaluate the limits
(1)$$\lim_{n\to\infty}\left[\sin\frac{\pi}{n}+\sin\frac{2\pi}{n}+\cdots+\sin\frac{n\pi}{n}\right]$$
(2)$$\lim_{n\to\infty}\left[\left(1+\frac1{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2\cdots{\left(1+\frac{n^2}{n^2}\right)}^n\right]^{\frac1{n}}$$
Thanks!

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In (1), did you mean to divide by $n$? –  robjohn Jan 18 '13 at 7:55
Both are direct applications of the concept of Riemann sums. But could you please avoid giving orders and, even more importantly, explain what you tried and why you are stuck? Thanks in advance. Otherwise you are just dumping your trash here... –  Did Jan 18 '13 at 7:55
In (2), didn't you mean a summation in the brackets? Or a full product? –  Babak S. Jan 18 '13 at 7:56
@did: Do they not both diverge to $\infty$ as written? –  robjohn Jan 18 '13 at 7:57
@rajkmalmath: You need to be very careful before you state a problem from a book here. I am seeing too many mistakes in transposition which cause some of us who love to solve such problems to solve the wrong problem. BTW I think, in the second problem, the exponent at the end should be $1/n^2$, not $1/n$. –  Ron Gordon Jan 18 '13 at 8:53

## 4 Answers

(1)

$$\sum_{r=1}^n \sin{\frac{r \pi}{n}} = \Im{\sum_{r=1}^n e^{i r \frac{\pi}{n}}}$$

$$= \Im{\left [ \frac{e^{i \frac{\pi}{n}} - e^{i \frac{(n+1)\pi}{n}}}{1-e^{i \frac{\pi}{n}}} \right ]}$$

$$= \Im{\left [ \frac{1 - e^{i \frac{n \pi}{n}}}{e^{-i \frac{\pi}{n}}-1} \right ]}$$

$$= \frac{2}{4 \sin^2{\frac{\pi}{2 n}}} \Im{\left [e^{i \frac{\pi}{n}}-1 \right ]}$$

$$= \frac{2}{4 \sin^2{\frac{\pi}{2 n}}} \sin{\frac{\pi}{n}}$$

$$= \cot{\frac{\pi}{2 n}}$$

The sum diverges as $n \rightarrow \infty$ because $\cot{\frac{\pi}{2 n}} \sim \frac{2 n}{\pi} (n \rightarrow \infty)$. That said, if the limit of interest were

$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \sin{\frac{r \pi}{n}}$$

then this limit would be $\frac{2}{\pi}$.

(2) Let $L$ be the proposed limit. Then

$$\log{L} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^n k \log{\left ( 1 + \frac{k^2}{n^2} \right )}$$

$$= \lim_{n \rightarrow \infty} n \int_0^1 dx \: x \log(1+x^2)$$

Obviously, this diverges. However, if the limit in question were

$$L = \lim_{n\to\infty}\left[\left(1+\frac1{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2\cdots{\left(1+\frac{n^2}{n^2}\right)}^n\right]^{\frac1{n^2}}$$

then, from above, we may easily deduce that

$$\log{L} = \int_0^1 dx \: x \log(1+x^2) = \log{2} - \frac{1}{2}$$

which you can easily show using integration by parts. Then

$$L = \frac{2}{\sqrt{e}}$$

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This $\log L$ does not correspond to (2) (you missed the exponents of every factor in the product). –  Did Jan 18 '13 at 8:29
Sigh, I really need to stop doing this at 3:30 AM... –  Ron Gordon Jan 18 '13 at 8:30
Maybe. Especially since answering this question did not seem so urgent, if you ask me. –  Did Jan 18 '13 at 8:32
I am a sucker for interesting-looking sums and integrals, I guess. –  Ron Gordon Jan 18 '13 at 8:37

‎Recall that‎

${‎‎‎‎\int_‎a^b}{f(x)} \ dx = \frac{b-a}{n}\lim‎_{n‎‎‎‎\to\infty‎}\sum_{‎i=0‎‎}‎^n f(a+i\frac{b-a}{n})‎‎‎‎$

then for (1) ‎we ‎have ‎‎$‎f(x)=\sin(\pi x) , ‎b=‎1‎ , ‎a=0$‎

There‎fore, we have ‎

‎‎$‎\lim_{n\to\infty}\sum_{‎i=0‎‎}‎^n \sin\frac{‎i ‎\pi‎}{‎n‎} =‎ ‎‎‎‎{\int_0^1}‎ ‎{sin \pi ‎x‎} ‎\ dx = ‎‎‎‎\frac{2}{‎\pi‎}‎‎$‎

Also, for (2) :

$f(x)=x \ln(1+x^2)$ , $b=1$ , $a=0$

$$\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac in\ln\left(1+(\frac{i}{n})^2\right)=\int_0^1x\ln(1+x^2)\,\mathrm{d}x =\frac{2}{\sqrt e}$$

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For the sum we may consider the first $\lfloor n/2\rfloor$ terms and then apply Jordan's inequality $$\frac{2}{\pi}x\le \sin x\le x, \space x\in[0,\pi/2]$$

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Using Riemann Sums:

Hint (1): $$\lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{k\pi}{n}\right)\frac1n=\int_0^1\sin(\pi x)\,\mathrm{d}x$$

Hint (2): $$\lim_{n\to\infty}\sum_{k=1}^n\frac kn\log\left(1+\frac{k^2}{n^2}\right)\frac1n=\int_0^1x\log(1+x^2)\,\mathrm{d}x$$

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