Prove that if the complex series $\sum_{\nu =1}^\infty u_\nu$ converges and the real sequence $(a_{\nu})_{\nu=1}^\infty$ is non-negative and decreasing, then $$\Big|\sum_{\nu =k+1}^\infty u_\nu a_\nu\Big|\leq a_{k+1}M_k,$$ where $$M_k=\sup_{n\geq k+1}\Big|\sum_{\nu=k+1}^n u_\nu\Big|.$$
I have been trying to prove this for a while using summation by parts, but I haven't been able to get very far. Could someone give me a hint?
Thanks to a helpful hint from @DanielFischer, I managed to figure it out.
By summation by parts, for $n\geq k+1$, we have $$\Biggl\lvert\sum_{\nu =k+1}^n u_\nu a_\nu\Biggr\rvert = \Biggl\lvert t_na_{n+1}-t_ka_{k+1}+\sum_{\nu=k+1}^n t_\nu(a_\nu-a_{\nu +1})\Biggr\rvert\,,$$ where $t_\nu=\sum_{\mu=1}^\nu u_\nu$. We write \begin{align} \sum_{\nu=k+1}^n t_\nu(a_\nu-a_{\nu +1}) &= \sum_{\nu=k+1}^n t_k(a_\nu-a_{\nu +1})+\sum_{\nu =k+1}^n (t_\nu -t_k)(a_\nu-a_{\nu +1}) \\ &=t_k(a_{k+1}-a_{n+1})+\sum_{\nu =k+1}^n (t_\nu -t_k)(a_\nu-a_{\nu +1}), \end{align} and since $a_\nu-a_{\nu+1}\geq 0$, we have \begin{align} \Biggl\lvert t_na_{n+1}-t_ka_{k+1}+\sum_{\nu=k+1}^n t_\nu(a_\nu-a_{\nu +1})\Biggr\rvert &= \Biggl\lvert (t_n-t_k)a_{n+1}+\sum_{\nu=k+1}^n(t_\nu-t_k)(a_\nu-a_{\nu+1})\Biggr\rvert \\ &\leq \lvert t_n-t_k\rvert a_{n+1} + \sum_{\nu=k+1}^n \lvert t_\nu-t_k\rvert (a_\nu-a_{\nu+1}). \end{align} Now, since $M_k=\sup_{n\geq k+1} \lvert t_n-t_k\rvert$, it follows that \begin{align} \lvert t_n-t_k\rvert a_{n+1} + \sum_{\nu=k+1}^n \lvert t_\nu-t_k\rvert (a_\nu-a_{\nu+1}) &\leq M_k\Biggl(a_{n+1}+\sum_{\nu=k+1}^n (a_\nu-a_{\nu +1})\Biggr) \\ &= M_k(a_{n+1}+a_{k+1}-a_{n+1}) \\ &= M_ka_{k+1}. \end{align}