Prove that for every $n \in \mathbb{N}$ $\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$

Question

$$\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$$

1. First step would be proving that the statement is true for n=2

On the LHS for $n=2$ we would have $\frac{1}{4}$ therefore the statement is true for $n=2$

2. Now we must assume the statement is true for $n=j$ with $j\geq2$

$$\sum\limits_{k=2}^{j}{\frac{1}{k^2}}<1$$

3. Now we must prove true for $n=j+1$

$\sum\limits_{k=2}^{j+1}{\frac{1}{k^2}}<1$

$\Rightarrow \sum\limits_{k=2}^{j}{\frac{1}{k^2}}+\frac{1}{(k+1)^2}<1$

I cant seem to introduce the induction hypothesis

Answer 1

Strengthen the induction hypothesis

Hint: Prove by induction that

$$ \sum_{k=2}^n \frac{1}{k^2 } < 1 - \frac{1}{n}.$$

Answer 2

Let $ n\in\mathbb{N} : $

Notice that for any integer $ k\geq 2 $, we have $ k^{2}\geq k\left(k-1\right) $, and thus : $$ \sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\sum_{k=2}^{n}{\frac{1}{k\left(k-1\right)}}=\sum_{k=2}^{n}{\left(\frac{1}{k-1}-\frac{1}{k}\right)}=1-\frac{1}{n}<1 $$

Answer 3

$f(x)= \frac{1}{x^2}$, $x >0$, stricly decreasing.

$ \displaystyle{\sum_{k=2}^{n}}\frac{1}{k^2} <\displaystyle{ \int_{1}^{n}}\frac{1}{x^2}dx =-\frac{1}{x}\big ]_1^n=$

$1-1/n. $

As suggested by Calvin you can use the above inequality for the induction proof.

Answer 4

Here is another approach :

We can easily prove that $ \left(\forall x\in\left[0,\frac{1}{2}\right]\right),\ \frac{2}{\pi}\arctan{\left(2x\right)}-x\geq 0 $ (By differentiation the function $ x\mapsto\frac{2}{\pi}\arctan{\left(2x\right)}-x $, studying its variations, etc...)

Thus, if $ n $ is a positive integer, we have the following : \begin{aligned}\sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{2}{k^{2}}\right)}}&=\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{\left(k+1\right)-\left(k-1\right)}{1+\left(k+1\right)\left(k-1\right)}\right)}}\\&=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{k}\right)}+\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{k}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\left(\arctan{\left(n+1\right)}-\arctan{2}\right)+\frac{2}{\pi}\left(\arctan{n}-\frac{\pi}{4}\right)\\ &\leq\frac{3}{2}-\frac{2}{\pi}\arctan{2}<1\end{aligned}

To get to the last line we upper bounded the $ \arctan $s by $ \frac{\pi}{2} \cdot $