Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction

Question

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!

Answer 1

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 $K$,

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

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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 :)

Answer 2

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 $J_0=\pi^3/24$.

Answer 3

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.

Answer 4

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).