# Does the limit of the sum equal the sum of the limits in this case?

I'm interested in evaluating the following limit: \begin{align*} \lim_{N\to\infty}\sum_{n= 1}^N\frac{1}{N\sin\left(\frac{\pi n}{N}\right)} \end{align*} Because \begin{align*} \lim_{N\to\infty}\frac{1}{N\sin\left(\frac{\pi n}{N}\right)} &= \frac{1}{\pi n} \end{align*} is the following reasoning justified? \begin{align*} \lim_{N\to\infty}\sum_{n = 1}^N\frac{1}{N\sin\left(\frac{\pi n}{N}\right)} &= \frac{1}{\pi} + \frac{1}{2\pi} + \frac{1}{3\pi} + \cdots + \frac{1}{\pi N} \\ &= \frac{1}{\pi}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{N}\right) \\ &= \text{DNE} \end{align*} In general, is there a more systematic way to evaluate such limits? I'm unaccustomed to seeing the upper limit of summation appear in the expression itself.

• Looks like a Riemann sum in disguise. – White Shirt Jan 5 '16 at 23:01
• To prove the limit is infinite, simply use $\sin(\pi x)\leqslant\pi x$ for every nonnegative $x$ and deduce that $$\sum_{n= 1}^N\frac{1}{N\sin\left(\frac{\pi n}{N}\right)}\geqslant\sum_{n= 1}^{N}\frac{1}{\pi n}\to\infty.$$ – Did Jan 5 '16 at 23:03
• @WhiteShirt Riemann sums are for continuous (hence bounded) functions. – Did Jan 5 '16 at 23:03
• Indeed, though OP's observation "seeing the upper limit of summation appear in the expression itself" reminded me of the formula $$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f\left (\frac{n}{N}\right )=\int_0^1 f\left (x\right ) \mathrm{d}x$$ In this case $f(x)=1/\sin(\pi x)$. Of course, the integral does not converge in $[0,1]$, so neither would the limit. – White Shirt Jan 5 '16 at 23:13