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since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $

my problem is can we prove the statement

"for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $"

only using induction ?(without using the fact that convergence of the above series)

any ideas, thanks!

Bad English
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    I'm not sure, but you can prove it for some rational numbers very close to $\dfrac {\pi^2}{6}$ – Ovi Feb 14 '18 at 13:36
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    @Ovi Due to the convergence that would not be enough as for any $r < \dfrac{\pi^2}6$, the finite sum for big enough $n$ will be $> r$. – orlp Feb 14 '18 at 13:39
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    Is there any geometric meaning of the number $\pi^2$? If yes, one could use that and an appropriate tiling. – Arnaud Mortier Feb 14 '18 at 13:40
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    @ArnaudMortier It's the area of the square such that you can roll a circle with diameter 1 around the perimeter of the square exactly 4 times. Or you can get $\pi^2$ as a line segment using this construction, getting $\pi$ once again by rolling a circle. – orlp Feb 14 '18 at 13:43
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    For what it's worth, I found this geometrical proof of the equality, one could potentially turn it into an induction. https://arxiv.org/pdf/math/0701039.pdf – Arnaud Mortier Feb 14 '18 at 13:52
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    @orlp I know, I wasn't suggesting it's enough – Ovi Feb 14 '18 at 13:54
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    @ArnaudMortier I guess one geometric meaning is that you can fill a square of area $\dfrac {\pi^2}{6}$ with squares of area $1, \dfrac {1}{2^2}$, $\dfrac {1}{3^2}, ...$ – Ovi Feb 14 '18 at 13:57
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    @Ovi Not necessarily. E.g. see a similar problem, which is open. – orlp Feb 14 '18 at 14:11
  • @orlp Ah yes, so it might not be possible. But if it is possible, it should be enough to show the equality. – Ovi Feb 14 '18 at 14:31

4 Answers4

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Take a look at this proof of Daniel J. Velleman. For $n\geq 1$, we define the positive numbers, $$I_n:=\int_0^{\pi/2}(\cos(x))^{2n}\,dx\quad\text{and}\quad J_n:=\int_0^{\pi/2}x^2(\cos(x))^{2n}\,dx.$$ By integration by parts we show that $$I_n=(2n-1)(I_{n-1}-I_n)\implies I_n=\frac{2n-1}{2n}I_{n-1}$$ and $$I_n=n(2n-1)J_{n-1}-2n^2J_n.$$ Hence, by dividing the last one by $n^2I_n$, we get $$\frac{1}{n^2}=\frac{(2n-1)J_{n-1}}{nI_n}-\frac{2J_n}{I_n}= 2\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right).$$ It follows by induction (see the telescopic sum) that for $K\geq 1$, $$\frac{\pi^2}{6}-\sum_{n=1}^K\frac{1}{n^2}=\frac{\pi^2}{6}-2\sum_{n=1}^K\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right)=\frac{\pi^2}{6}-2 \left(\frac{J_0}{I_0}-\frac{J_K}{I_K}\right)=\frac{2J_K}{I_K}>0$$ where we used that $I_0=\pi/2$ and $J_0=\pi^3/24$.

Robert Z
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This is not a full proof yet, just an idea I had.

From Leibniz series we know that for $k \geq 1$:

$$\frac{\pi}{4} \geq 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1}$$

Squaring we have:

$$\frac{\pi^2}{6} \geq \frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2$$

If we could prove that for any $k \geq 1$:

$$\frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2 \geq 1+\frac{1}{2^2}+\dots+\frac{1}{k^2}$$

We are finished.

The base case:

$$\frac{8}{3} \left( 1-\frac{1}{3}\right)^2=\frac{32}{27} >1$$

The induction hypothesis is:

$$\frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2 \geq 1+\frac{1}{2^2}+\dots+\frac{1}{k^2} \tag{1}$$

We need to prove that from (1) follows:

$$\frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k+1}-\frac{1}{4k+3} \right)^2 \geq 1+\frac{1}{2^2}+\dots+\frac{1}{k^2}+\frac{1}{(k+1)^2} \tag{2}$$

If this is indeed true, I'll try to finish the proof. Or I offer anyone else who wants to try the opportunity to post their answer before me.

For a few values of $k$ I checked the inequality (1) holds, so it should be possible to prove it by induction.

Yuriy S
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The statement that we will prove with induction is that for every $K > k^*$,

$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - \frac 1K $$ where $k^* = 4091641$.

This implies that, for every $K$,

$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 $$

The idea is to find a function $f(K) > 0$ such that

$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - f(K) $$

As @orlp states in his comment, it is clear that $f(K)$ cannot be a constant, because we know that the partial sums get arbitrarily close to $\frac{\pi^2}6$. So which function do we pick?

Well, just write the inductive step:

$$\sum_{n=1}^K \frac {1}{n^2} + \frac{1}{(K+1)^2} \le \frac {\pi ^2} 6 - f(K) + \frac{1}{(K+1)^2} \mathop{\le}^? \frac {\pi ^2} 6 - f(K+1)$$

If we can prove the last inequality, then the inductive step will work and we will have solved the problem.

The last inequality is equivalent to

$$f(K+1) \le f(K) - \frac 1{(K+1)^2}$$

which means that $f(K)$ must be decreasing. One can easily check that $f(K) = \frac 1K$, for example, works. The only thing that is left is the base case.

Turns out, the base case is not so easy to find. A quick numerical simulation, though, proved that for $n=4091641$ the case case is satisfied, and the rest follows. Note that since we know that the statement is true for a certain $K > k^*$, we also know that every partial sum up to $k^*$ is $\le \frac{\pi^2}6$, as the partial sums are increasing.

One could probably find a better $f(K)$ such that we don't need computers to verify the base case, but I'll leave that to someone else :)

Ant
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  • Could you please consider leaving your answer up even when you find an error? I'm interested in it but I can't spend much time on stackexchange for a few days. – Ovi Feb 14 '18 at 14:33
  • @Winther Very good point, I was mistaken. A base case exists though; around $n=4091641$. It feels a bit like cheating though :D – Ant Feb 14 '18 at 14:34
  • @Ovi Sure. In any case the mistake was in not checking the base case correctly. A base case still exists, but needs to be found with computers :-) – Ant Feb 14 '18 at 14:39
  • The proof cannot be right because if you replace $\pi^2/6$ by $\pi^2/6-10^{-50}$ (second term chosen so that the base case holds), the arguments still holds but leads to a wrong conclusion. –  Feb 14 '18 at 15:08
  • @YvesDaoust Good point.. I was thinking there was something suspicious about this. But where is the mistake? – Ant Feb 14 '18 at 15:14
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    Your final inequality amounts to $\sum_{n=1}^K\frac1{n^2}+f(K) $ being decreasing, and since we know this converges to $\pi^2/6$, it actually proves the convergence is from above. – Vincenzo Oliva Feb 14 '18 at 16:30
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    In fact, for the same reason, we would like $\sum_{n=1}^K\frac1{n^2}+f(K)$ to be increasing. And it obviously works with $f(K)=\frac1{(K+1)^2}$. But then again, not only this hardly differs from $f(K)=0$, regardless it heavily uses the convergence to $\pi^2/6$, which the OP didn't want to consider. – Vincenzo Oliva Feb 14 '18 at 16:43
  • @VincenzoOliva Thanks for your comment, but I am confused. How does my inequality amount to that object being decreasing? Also, I don't get what works with $f(K) = \frac 1{(K+1)^2}$. Finally, where did I use the convergence of the sum? I have only considered partial sums – Ant Feb 14 '18 at 17:05
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This is unlikely. The bare induction step would be

$$S_n\le\frac{\pi^2}6\implies S_n+\frac1{n^2}\le\frac{\pi^2}6,$$ which obviously doesn't hold. There is not enough information in the inductive hypothesis.

Any information on the asymptotics of the series will be of the form $\dfrac{\pi^2}6-\epsilon(n)$, which contains "the answer" (i.e. a hint on convergence).