Convergence of $\lim_{N\to\infty}\sum_{n=1}^N \exp(-N\sin^2(\frac{n\pi}{2N}))$ and $\lim_{N\to\infty}\sum_{n=1}^N \exp(-\sin^2(\frac{n\pi}{2N}))$ I can't find the right approach to tackle the question whether
$$\lim_{N\to\infty} \sum_{n=1}^N \exp\Bigg(-N \sin^2\left(\frac{n\pi}{2N}\right)\Bigg)$$
and
$$\lim_{N\to\infty} \sum_{n=1}^N \exp\Biggl(-\sin^2\left(\frac{n\pi}{2N}\right)\Biggr)$$
converge or diverge. The fact that the limiting variable appears both as the upper bound of summation as well as in the individual summands seems to make the standard methods known to me inapplicable.
I suspect that the second limit (i.e. the one not containing $N$ in the exponent directly) does not exist, but that the first one may. I would be very grateful if you could point me to methods that allow one to determine the existence of the limits.
If a limit exists, I would also be very interested in understanding how, if at all, one could (approximately) replace the sum with an integral.
(For background, these questions have arisen during my study of the Rouse theory of polymer dynamics, e.g. in chapter 7.3.2. of Doi and Edwards, "The Theory of Polymer Dynamics". Physical explanations of how one can justify the treatment therein would be very welcome, too.)
Thank you in advance!
 A: Both sums diverge and a general idea to prove it is to use Riemann sums, just as suggested in the comments. The second sum was proved divergent in the comments. For the first sum notice that $0 \le \sin x \le x$ for $x \in [0, \pi/2]$ implies $0 \geq - \sin^2 x \geq -x^2$ in this range and since the exponential function is increasing we have
\begin{align*}
\sum_{n=1}^N \exp\Bigg(-N \sin^2\left(\frac{n\pi}{2N}\right)\Bigg)
  &\geq \sum_{n=1}^N \exp\Bigg(- \frac{n^2\pi^2}{4N}\Bigg) \\
  &= \sqrt{N} \cdot \left[ \frac{1}{\sqrt{N} } \sum_{n=1}^N \exp\Bigg(- \frac{\pi^2}{4} \bigg(\frac{n}{\sqrt{N}}\bigg)^2 \Bigg)\right]
\end{align*}
Now, the expression inside square brackets is close to the integral (you can compare them using integral test for instance):
$$
  \int_0^{\sqrt{N}} e^{-\frac{\pi^2}{4} x^2} dx \to \int_0^\infty e^{-\frac{\pi^2}{4} x^2} dx < \infty \quad \text{as $N \to \infty$}.
$$
This implies the first integral grows at least as a multiple of $\sqrt{N}$ and must diverge.
A: Here's another way to see the limit of the first series is $\infty.$ Verify that for each fixed $n,$
$$\lim_{N\to \infty}\exp\left(-N \sin^2\left(\frac{n\pi}{2N}\right)\right) = 1.$$
Fix $N_0\in \mathbb N.$ Then for $N\ge N_0,$ the first series is at least
$$\sum_{n=1}^{N_0}\exp\left(-N \sin^2\left(\frac{n\pi}{2N}\right)\right).$$
This is a finite sum, so computing its limit as $N\to \infty$ is easy: we get $\sum_{n=1}^{N_0}1= N_0.$ Since $N_0$ is arbitrarily large, the limit of the first series is $\infty.$ Since the terms of the second series are at least as large as the those of the first series, the limit for the second series is also $\infty.$
