Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / D / D3

The monotone convergence theorem states that if nonnegative measurable functions $h_n$ increase almost everywhere to $h$, then $\int h={\lim }_n\int h_n$, allowing infinity. The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ for an integrable $g$, then $h$ is integrable and $\int h_n\to \int h$.

Apply monotone convergence to $A_n={\sum }_{k=1}^n|a_k|$. Its limit $A={\sum }_{k\ge 1}|a_k|$ has integral ${\sum }_k\int |a_k|<\infty$, so $A<\infty$ almost everywhere. Thus the original series converges absolutely almost everywhere. Its partial sums are bounded in absolute value by the integrable function $A$, and their limit $s$ satisfies $|s|\le A$. Dominated convergence givesAssigning any value on the null exceptional set defines an integrable representative of $s$.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. All terms after multiplication by $(\log x{)}^2$ are nonnegative, so monotone convergence permits integration term by term. Substituting $x=e^{-t}$ and integrating by parts twice givesConsequently the integral is ${\sum }_{k\ge 1}2/k^2$, and the supplied sum gives $\boxed{{\pi }^2/3}$.