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

Question

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!

Answer 1

(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

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

Answer 2

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

Answer 3

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

Answer 4

‎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) :

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

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