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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?

Page David
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2 Answers2

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Hint:

$$\begin{align} &1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9}+\frac{1}{10^2}+\cdots \\&=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{10^2}+\cdots\\&+1+ \frac14+\frac19+\frac1{16}+\frac1{25}+\cdots \\&<\sum\frac 1{n^2} + \sum\frac1{n^2} \end{align}$$

player3236
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Since every positive integer has exactly one perfect square, we can write

\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*}

It follows that the series converges.

Alann Rosas
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