Solution
= Solution
\b[True.] For $m>n$, $|a_m-a_n|\le\sum_{j=n}^{m-1}d_j$. Convergence of $\sum d_j$ makes this tail arbitrarily small, so $(a_n)$ is Cauchy and hence convergent.
= Solution
\b[True.] For $m>n$, $|a_m-a_n|\le\sum_{j=n}^{m-1}d_j$. Convergence of $\sum d_j$ makes this tail arbitrarily small, so $(a_n)$ is Cauchy and hence convergent.