Finding the positive integer numbers to get $\frac{\pi ^2}{9}$

From $$\frac{\pi^2}8=\sum_{2\nmid n}\frac1{n^2}$$ we find that $$\frac{\pi^2}{72}=\frac1{3^2}\sum_{2\nmid n}\frac1{n^2}=\sum_{2\nmid n, 3\mid n}\frac1{n^2}$$ and hence $$\begin{align}\frac{\pi^2}9=\frac{\pi^2}8-\frac{\pi^2}{72}&=\sum_{2\nmid n, 3\nmid n}\frac1{n^2}\\&=\frac1{1^2}+\frac1{5^2}+\frac1{7^2}+\frac1{11^2}+\frac1{13^2}+\frac1{17^2}+\frac1{19^2}+\frac1{23^2}+\frac1{25^2}+\ldots\end{align}$$

You don't actually need to know that $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$ to get such a sequence. Instead, you can use this theorem (which is probably an exercise in some early chapter of Rudin):

Theorem: Let $\sum_k s_k$ be a convergent sum of positive numbers, say $S=\sum_k s_k$. Suppose that, for all $n$, we have $s_n < \sum_{k=n+1}^\infty s_k$. Then there is a subsequence of $s_k$ whose sum converges to $x$ for any positive $x < S$.

Proof: Construct the subsequence greedily. That is, given some partial sum $s_{k_1}+\dots+s_{k_{i-1}}$, let $k_i$ be the least integer greater than $k_{i-1}$ with $s_{k_1}+\dots+s_{k_i} < x$. As the $s_k$ become arbitrarily small, such a $k_i$ will always exist.

By construction, the partial sums of $\sum s_{k_i}$ are bounded above by $x$; as the $s_k$ are all positive, it follows that $\sum s_{k_i}$ converges to some number which is at most $x$.

Now, we will need the following lemma:

Lemma: The subsequence $s_{k_i}$ omits infinitely many elements of $s_k$.

Proof: As $0<x<S$, we can find $n_1$ such that $\sum_{k=1}^{n-1} s_k< x < \sum_{k=1}^{n_1} s_k$. Then by construction $s_{k_i}$ will include $s_1,\dots,s_{n_1-1}$ but not $s_{n_1}$, and so $s_{k_i}$ omits the element $s_n$. By hypothesis, $\sum_{k=n+1}^\infty s_k > s_n$, and so $$x-\sum_{k=1}^{{n_1}-1} s_k < s_{n_1} < \sum_{k=n_1+1}^\infty s_k$$ But then the same argument gives us some $n_2>n_1$ with $$ \sum_{k=n_1+1}^{n_2-1} s_k < x-\sum_{k=1}^{n_1-1} s_k < \sum_{k=n_1+1}^{n_2} s_k $$ and similarly $s_{k_i}$ will omit $s_{n_2}$. Proceeding in this manner, we obtain an infinite sequence $s_{n_j}$ of omitted terms. $\square$

With this lemma, the theorem is straightforward. Choose $\epsilon >0$. As there are infinitely many omitted $s_{n_j}$ and the sequence $s_k \to 0$, we can find $j$ so that $s_{n_j} < \epsilon$ is omitted. Now, take $i$ so that $s_{k_i}$ is the latest term in the subsequence prior to $s_{n_j}$. Then, as $s_{n_j}$ is omitted, $s_{k_1}+\dots+s_{k_i} + s_{n_j} > x$. It follows that $s_{k_1}+\dots+s_{k_i} > x-\epsilon$. As we have already shown that partial sums of $\sum_i s_{k_i}$ are all bounded above by $x$, this completes the proof that $x=\sum_{i=1}^\infty s_{k_i}$. $\blacksquare$

Now, as $\frac{1}{n^2}$ is a decreasing sequence, we can use the integral test to bound the tails of the sum $\sum \frac{1}{n^2}$: $$ \sum_n^\infty \frac{1}{k^2} \geq \int_n^\infty \frac{1}{x^2} \, dx=\frac{1}{n} \, . $$ If $n \geq 2$, $\frac{1}{n+1} > \frac{1}{n^2}$, and so the theorem holds — for the series $\frac{1}{2^2}+\frac{1}{3^2}+\dots$. But $\frac{\pi^2}{9}>1$, so that's good enough; we just apply the theorem to $x=\frac{\pi^2}{9} - 1$.

In fact, if we now allow ourselves to know that $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$ again, this method implies that $x$ can be written as an infinite sum of the reciprocals of distinct squares whenever $x \in \left(0,\frac{\pi^2}{6}-1\right] \cup \left(1, \frac{\pi^2}{6}\right]$.

$$\frac{\pi ^2}{9}=\frac{\pi ^2}{6}\times 6 \times \frac{1}{9}=(\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+ \cdots)\times 6$$