# Infinite Sums which turn out to be Riemann Integrals

I'm looking for examples of infinite series which look hard to evaluate at first, but become very simple when viewed as a Riemann integral. An example would be $$\frac{1}{n+1}+\frac{1}{n+2}+ \ldots +\frac{1}{2n}=\sum_{i=1}^{n}\frac{1}{n+i}\\=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+i/n} \to \int_{0}^{1}\frac{1}{1+x}dx=\log 2$$ (I just think they're cute)

• I really want to help, but ... if we take any integrable function $f$ on $[a,b]$, and have a look at its Riemann sum $$\lim_{n\to \infty}\frac{b-a}{n}\sum_{\ell=1}^nf\left(a+\ell\frac{b-a}{n}\right)$$ then, could we not view any of these for which we can evaluate $\int_a^bf(x)dx$ in closed as "cute?" What specifically are you seeking here? – Mark Viola Jul 10 '15 at 23:09
• What I'm looking for is an example where using a Riemann sum is surprising - I've seen examples of it before and unfortunately nothing springs to mind, but I think there are times when summing a series as a Riemann integral leads to a really slick solution. I'll admit it's a slightly vague question. – preferred_anon Jul 10 '15 at 23:12