I have the series $\sum a_n$ where
$$a_n = \begin{cases} \frac{1}{n}, &\text{$n$ is a perfect square} \\ \frac{1}{n^2}, &\text{otherwise}\end{cases}$$
Prove that $\sum_{n=1}^\infty R_n$ converges where remainder $R_n = \sum_{i=n}^\infty a_i$.
My work: I can show that the series $\sum a_n$ converges because
$$\sum_{n=1}^Na_n = \sum_{n \neq k^2} \frac{1}{n^2} + \sum_{n = k^2, 1 \leq k^2 \leq N} \frac{1}{k^2} \leq 2 \sum_{n=1}^N \frac{1}{n^2}$$
and this means the remainders converge to $0$, $\lim_{n\to \infty}R_n = 0$. Hence, I can't rule out convergence of $\sum R_n$ by the term divergence test. But I can't find a comparison to prove it converges.