IMC 1996 · Day 2 · P11

(i) Set $f(t) = \dfrac{t}{(1 + t^2)^2}$, $h = \dfrac{1}{\sqrt{x}}$. Then $$\sum_{n=1}^{\infty} \frac{nx}{(n^2 + x)^2} = h \sum_{n=1}^{\infty} f(nh) \xrightarrow[h \to 0]{} \int_0^{\infty} f(t)\,dt = \frac{1}{2}.$$ The convergence holds since $h \sum\limits_{n=1}^{\infty} f(nh)$ is a Riemann sum of the integral $\int\limits_0^{\infty} f(t)\,dt$. There are no problems with the infinite domain because $f$ is integrable and $f \downarrow 0$ for $x \to \infty$ (thus $h \sum\limits_{n=N}^{\infty} f(nh) \ge \int\limits_{Nh}^{\infty} f(t)\,dt \ge h \sum\limits_{n=N+1}^{\infty} f(nh)$).

(ii) We have $$\tag{1} \left| \sum_{n=1}^{\infty} \frac{nx}{(n^2+x)^2} - \frac{1}{2} \right| = \left| \sum_{n=1}^{\infty} \left( h f(nh) - \int_{nh - \frac{h}{2}}^{nh + \frac{h}{2}} f(t)\,dt \right) - \int_0^{\frac{h}{2}} f(t)\,dt \right| \le \sum_{n=1}^{\infty} \left| h f(nh) - \int_{nh - \frac{h}{2}}^{nh + \frac{h}{2}} f(t)\,dt \right| + \int_0^{\frac{h}{2}} f(t)\,dt$$ Using twice integration by parts one has $$\tag{2} 2b\,g(a) - \int_{a-b}^{a+b} g(t)\,dt = -\frac{1}{2} \int_0^b (b-t)^2 (g''(a+t) + g''(a-t))\,dt$$ for every $g \in C^2[a-b, a+b]$. Using $f(0) = 0$, $f \in C^2[0, h/2]$ one gets $$\tag{3} \int_0^{h/2} f(t)\,dt = O(h^2).$$ From (1), (2) and (3) we get $$\left| \sum_{n=1}^{\infty} \frac{nx}{(n^2+x)^2} - \frac{1}{2} \right| \le \sum_{n=1}^{\infty} h^2 \int_{nh - \frac{h}{2}}^{nh + \frac{h}{2}} |f''(t)|\,dt + O(h^2) = h^2 \int_{\frac{h}{2}}^{\infty} |f''(t)|\,dt + O(h^2) = O(h^2) = O(x^{-1}).$$