= Solution
Since $a_j=S_j-S_{j-1}$, shifting an index yields the <summation by parts> identity
$$
\begin{aligned}
\sum_{j=m}^n a_jb_j&=\sum_{j=m}^nS_jb_j-\sum_{j=m}^nS_{j-1}b_j\\
&=S_nb_n-S_{m-1}b_m+\sum_{j=m}^{n-1}S_j(b_j-b_{j+1}).
\end{aligned}
$$
The sum on the last line is empty when $m=n$, so the identity also covers that case.
\b[The series with a bounded monotone multiplier converges.] Here is a proof that also permits conditional convergence. Choose $M>0$ with $|b_j|\leq M$. 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:
$$
\sum_{j=m}^n a_jb_j=T_nb_n+\sum_{j=m}^{n-1}T_j(b_j-b_{j+1}).
$$
Since the multiplier is a <monotone sequence>, $\sum_{j=m}^{n-1}|b_j-b_{j+1}|=|b_m-b_n|$. Consequently
$$
\left|\sum_{j=m}^n a_jb_j\right|\leq\eta\bigl(|b_n|+|b_m-b_n|\bigr)\leq3M\eta.
$$
Taking $\eta=\varepsilon/(3M)$ 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>.
\b[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 $h(x)=\log x/x$,
$$
h'(x)=\frac{1-\log x}{x^2}<0\qquad(x\geq3).
$$
Thus $n^{1/n}=\exp(h(n))$ decreases for $n\geq3$; it is bounded and tends to $1$, because $\log n/n\to0$. Apply the proved multiplier result to the tail beginning at $3$. The fact that the first few values are not monotone causes no difficulty.
Back to article page
