# Convergence of this sum?

A very simple example from my textbook $$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9}+\frac{1}{10^2}+\cdots$$ or say $$a_n=1/n^2$$ if $$n$$ is not a perfect square, otherwise $$a_n=1/n$$. The book simply says by comparing this sum with $$\Sigma1/n^2$$ will show it converges.

But I think because $$1/n^2\le a_n$$, I cannot conclude the convergence of $$\Sigma a_n$$ by the convergence of $$\Sigma1/n^2$$. Could you show me the detail about why it converges?

• When you say "perfect number", do you mean "perfect square"? Sep 15, 2020 at 5:52
• Notice though that $\frac{1}{4}=\frac{1}{2^2}$, $\frac{1}{9}=\frac{1}{3^2}$, etc. So you can use this to simplify the sum above. Sep 15, 2020 at 6:20

\begin{align} &1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9}+\frac{1}{10^2}+\cdots \\&=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{10^2}+\cdots\\&+1+ \frac14+\frac19+\frac1{16}+\frac1{25}+\cdots \\&<\sum\frac 1{n^2} + \sum\frac1{n^2} \end{align}
\begin{align*} & 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9}+\cdots+\frac{1}{15^2}+\frac{1}{16}+\frac{1}{17^2}+\cdots\\ &= 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{3^2}+\cdots+\frac{1}{15^2}+\frac{1}{4^2}+\frac{1}{17^2}+\cdots\\ &= 1+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\left(\frac{1}{3^2}+\frac{1}{3^2}\right)+\left(\frac{1}{4^2}+\frac{1}{4^2}\right)+\cdots\\ &= 1+\frac{2}{2^2}+\frac{2}{3^2}+\frac{2}{4^2}+\cdots\\ &< 2+\frac{2}{2^2}+\frac{2}{3^2}+\frac{2}{4^2}+\cdots\\ &= 2\sum_{n=1}^{\infty}\frac{1}{n^2} \end{align*}