Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 9F (Analysis I)

Since $a_j=S_j-S_{j-1}$, shifting an index yields the summation by parts identityThe sum on the last line is empty when $m=n$, so the identity also covers that case.

The series with a bounded monotone multiplier converges. Here is a proof that also permits conditional convergence. Choose $M>0$ with $|b_j|\le M$. The Cauchy sequence criterion for the convergent series $\sum a_j$ ensures that, for any $\eta >0$ and sufficiently large $m$, all the partial tails $T_j={\sum }_{k=m}^j{a}_k$ satisfy $|T_j|<\eta$. Apply summation by parts to these partial tails:Since the multiplier is a monotone sequence, ${\sum }_{j=m}^{n-1}|b_j-b_{j+1}|=|b_m-b_n|$. ConsequentlyTaking $\eta =\epsilon /(3M)$ proves that the weighted partial sums form a Cauchy sequence, and hence converge. This is a special case of bounded-variation multipliers of a convergent series.

The series $\sum n^{1/n}{a}_n$ also converges. The multiplier need only be monotone eventually, since a finite initial segment cannot affect convergence. For $h(x)=\log x/x$,Thus $n^{1/n}=\exp (h(n))$ decreases for $n\ge 3$; it is bounded and tends to $1$, because $\log n/n\to 0$. Apply the proved multiplier result to the tail beginning at $3$. The fact that the first few values are not monotone causes no difficulty.