How to prove $\sum_p p^{-2} < \frac{1}{2}$?

Question

I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me.

Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of squared prime reciprocals) deal with the exact value of the above sum, and so require some non-elementary math.

Answer 1

All primes but 2 are odd numbers so $$\sum_p p^{-2} < 1/4 + \sum_{k=1}^\infty \frac{1}{(2k+1)^2}$$ Using the fact that $1/x^2$ is convex the sum is bounded by $$ \sum_{k=1}^\infty \int_{k-1/2}^{k+1/2}\frac{1}{(2x+1)^2}dx = \int_{1/2}^\infty \frac{1}{(2x+1)^2}dx = 1/4$$

Answer 2

If you know $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ then you could simply say $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} $$ $$\lt \frac{\pi^2}{6} - \frac{1}{1^2}- \frac{1}{4^2}- \frac{1}{6^2}- \frac{1}{8^2}- \frac{1}{9^2}- \frac{1}{10^2}- \frac{1}{12^2}- \frac{1}{14^2}- \frac{1}{15^2}- \frac{1}{16^2} $$ $$ \approx 0.49629 $$ $$ \lt \frac12.$$

Alternatively if you do not know that, instead use $\displaystyle \frac{1}{k^2} \le \int_{x=k-1}^k \frac{1}{x^2}\, dx = \frac{1}{k-1} - \frac{1}{k}$ so $\displaystyle \sum_{n=k}^\infty \frac{1}{n^2} \le \int_{x=k-1}^\infty \frac{1}{x^2}\, dx = \frac{1}{k-1}$ and you can say: $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} \lt \frac{1}{2^2}+ \frac{1}{3^2}+ \frac{1}{5^2}+ \frac{1}{7^2}+ \frac{1}{11^2}+ \frac{1}{13^2}+ \frac{1}{17-1} \approx 0.4982 \lt \frac12.$$

Answer 3

We can deduce this quickly, and without knowing the numerical value of $\pi$, from the fact that $$\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6},$$ for which there are numerous proofs available.

Let $E$ denote the set of even numbers; the sum of the squares of all such numbers is $$\sum_{n \in E} \frac{1}{n^2} = \sum_{k \in \Bbb N} \frac{1}{(2 k)^2} = \frac{1}{4} \sum_{k \in \Bbb N} \frac{1}{k^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24}.$$ Now, let $X$ denote the union of $\{2\}$ and all positive odd integers $> 1$. In particular, $X$ contains the set $\Bbb P$ of all prime numbers as a subset, and so \begin{align} \sum_{p \in \Bbb P} \frac{1}{p^2} &\leq \sum_{n \in X} \frac{1}{n^2} \\ &= \sum_{n \in \Bbb N} \frac{1}{n^2} - \sum_{n \in E} \frac{1}{n^2} - \frac{1}{1^2} + \frac{1}{2^2} \\ &= \frac{\pi^2}{6} - \frac{\pi^2}{24} - 1 + \frac{1}{4} \\ &= \frac{\pi^2}{8} - \frac{3}{4} . \end{align} So, it suffices to show that $$\frac{\pi^2}{8} - \frac{3}{4} < \frac{1}{2},$$ but rearranging shows that this is equivalent to $\pi^2 < 10$, and $\pi < \frac{22}{7}$ implies $$\pi^2 < \left(\frac{22}{7}\right)^2 = \frac{484}{49} < \frac{490}{49} = 10. $$

Answer 4

Using easy inequalities and a telescoping sum along the way, we have

$$\begin{align} \sum_p{1\over p^2} &={1\over4}+{1\over9}+{1\over25}+{1\over7^2}+{1\over11^2}+{1\over13^2}+\cdots\\ &\lt{1\over4}+{1\over9}+{1\over25}+{1\over6\cdot8}+{1\over10\cdot12}+{1\over12\cdot14}+\cdots\\ &\lt{1\over4}+{1\over9}+{1\over25}+{1\over6\cdot8}+{1\over8\cdot10}+{1\over10\cdot12}+\cdots\\ &={1\over4}+{1\over9}+{1\over25}+{1\over2}\left(\left({1\over6}-{1\over8}\right)+\left({1\over8}-{1\over10}\right)+\left({1\over10}-{1\over12}\right)+\cdots\right)\\ &={1\over4}+{1\over9}+{1\over25}+{1\over12}\\ &\lt{1\over4}+{1\over8}+{1\over24}+{1\over12}\\ &={1\over2} \end{align}$$

Answer 5

This requires a little bit of creativity, but it works. First render by numerical calculation

$\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{25}+...+\dfrac{1}{289}<0.440$

Now for the rest we have

$\displaystyle {\sum_{p\text{ prime}, \\\ p \ge 19}\dfrac{1}{p^2}<\sum_{p\ge 19}\dfrac{1}{p(p-1)}}$

and use the relationship

$\dfrac{1}{p(p-1)}=\dfrac{1}{p-1}-\dfrac{1}{p}$

to telescope the last sum to $1/18<0.056$. Thereby

$\displaystyle \sum_{p\text{ prime}, \\\ p\ge 2}\dfrac{1}{p^2}<0.440+0.056=0.496$.

Answer 6

Here's a solution that exploits a comment of Oscar Lanzi under my other answer (using an observation that I learned from a note of Noam Elkies [pdf]). In particular, it avoids both the identity $\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6}$ and using integration.

Let $\Bbb P$ denote the set of prime numbers and $X$ the union of $\{2\}$ and the set of odd integers $> 1$; in particular $\Bbb P \subset X$, so where $E$ denotes the set of positive, even integers: $$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac{1}{n^2} = \color{#00af00}{\sum_{n \in \Bbb N \setminus E} \frac{1}{n^2}} - \frac{1}{1^2} + \frac{1}{2^2}.$$

Now, $$\sum_{n \in \Bbb N} \frac{1}{n^2} < 1 + \sum_{n \in \Bbb N \setminus \{1\}} \frac{1}{n^2 - \frac{1}{4}} = 1 + \sum_{n \in \Bbb N \setminus \{1\}} \left(\frac{1}{n - \frac{1}{2}} - \frac{1}{n + \frac{1}{2}} \right) = 1 + \frac{2}{3} = \frac{5}{3};$$ the second-to-last equality follows from the telescoping of the sum in the third expression.

The sum over just the even terms satisfies $$\sum_{m \in E} \frac{1}{m^2} = \sum_{n \in \Bbb N} \frac{1}{(2 n)^2} = \frac{1}{4} \sum_{n \in \Bbb N} \frac{1}{n^2} ,$$ and thus $$\color{#00af00}{\sum_{n \in \Bbb N \setminus E} \frac{1}{n^2} = \left(1 - \frac{1}{4}\right) \sum_{n \in \Bbb N} \frac{1}{n^2} < \frac{3}{4} \cdot \frac{5}{3} = \frac{5}{4}}.$$

Substituting in the first display equation above yields $$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac {1}{n^2} < \color{#00af00}{\frac{5}{4}} - 1 + \frac{1}{4} = \frac{1}{2} .$$