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

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.

• Shouldn't something like this work? $$\sum\limits_p\frac{1}{p^2} < \sum\limits_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2=\frac{\pi\cdot \pi}{24}<\frac12?$$ Nov 29, 2015 at 21:42
• Depending on how you define elementary, it's easy enough to just compare $\sum\limits_pp^{-2}$ to $\sum\limits_nn^{-2}=\frac{\pi^2}{6}$. In particular $$\pi^2/6-\sum\limits_{c\text{ nonprime}, ~c=1}^{16}<0.5$$ Nov 29, 2015 at 21:42
• I was looking for something that doesn't use $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$. Am I being too greedy? Nov 29, 2015 at 21:46
• @Ale: $\dfrac{\pi^2}{24} \approx 0.41123$ which is lower than the sum which is about $0.45224742$ Nov 29, 2015 at 21:48
• Yes. I think Ale's sum doesn't have $1/9$. Nov 29, 2015 at 21:49

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 $$\sum_{k=1}^\infty \int_{k-1/2}^{k+1/2}\frac{1}{(2x+1)^2}dx = \int_{1/2}^\infty \frac{1}{(2x+1)^2}dx = 1/4$$

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: $$\displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} \lt \frac{1}{2^2}+ \frac{1}{3^2}+ \frac{1}{5^2}+ \frac{1}{7^2}+ \frac{1}{11^2}+ \frac{1}{13^2}+ \frac{1}{17-1} \approx 0.4982 \lt \frac12.$$

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 $$\pi^2 < \left(\frac{22}{7}\right)^2 = \frac{484}{49} < \frac{490}{49} = 10.$$

• You can easily render $\pi^2<10$ directly from the Basel sum. First subtract off the $1/1^2$ term. For the rest render $n^2>(n+(1/2))(n-(1/2))$ so $(1/2^2)+(1/3^2)+(1/4^2)+...<(1/((3/2)×(5/2)))+(1/((5/2)×(7/2)))+(1/((7/2)×(9/2)))+...$. Then use the relationship $1/(x(x-1))=(1/(x-1))-(1/x)$ to telescope the last sum to $2/3$. Jan 5, 2020 at 16:28
• Thanks for pointing this out, Oscar. I first saw essentially this observation in a note of Noam Elkies: people.math.harvard.edu/~elkies/Misc/pi10.pdf Jan 5, 2020 at 19:38
• If you extract more terms you can sharpen this bound. Extracting four terms is sufficient to prove $\pi<22/7$. Jan 5, 2020 at 19:43
• @OscarLanzi I've used your observation to write up another answer that avoids the Basel sum altogether. Jan 5, 2020 at 23:25

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}

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

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

• OK notation gurus, I want the $p$ prime stuff under the summation sign not a subscript. My feeble attempt failed as the world did not like \Sum, and using \Sigma gave me a subscript. Need a solution, thanks. Jan 4, 2020 at 0:24
• Oscar, I tried to improve your latex. (for instance by replacing Sigma by sum). If you don't like it just roll back... There is also another possibility which maintains the Sigma using underset{} - but this seemed to me an overkill... Jan 4, 2020 at 8:47
• Gottfried used \displaystyle, which changes the behavior of how \sum interacts with _ and ^. But if you are in inline math mode and want this behavior without invoking \displaystyle, use \limits. Like \sum\limits_{p \text{ prime}}. Jan 4, 2020 at 9:15

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 $$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac {1}{n^2} < \color{#00af00}{\frac{5}{4}} - 1 + \frac{1}{4} = \frac{1}{2} .$$

• Aha, you used the comparison I set up, not the sum value. Good way to develop my "invention". Uh, what is that color command? Are you using three hex numbers for red, green and blue? (So that 0000ff for instance would be just blue?) Jan 5, 2020 at 23:34
• Yes, exactly, and it gives an improvement over my other solution, I think---thanks again! The color command can be used as in \color{#00af00}{This text is green.}. Jan 5, 2020 at 23:43