4

Question:

Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}, \text{ for all natural } n$$

My attempt:

Base Case: $n=1$ is true:

I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k},$ for some natural $k.$

Now we prove true for $n = k+1$

$$ 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2},\text{ by induction hypothesis} $$

Now how do I show that $2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{\left(k+1\right)}\text{ ?}$

Have I done everything correctly up until here?

If yes, how do I show this inequality is true?

Any help would be appreciated.

Michael Rozenberg
  • 194,933

2 Answers2

3

You are right!

We need to prove that: $$\frac{1}{(k+1)^2}<\frac{1}{k}-\frac{1}{k+1}$$ or $$\frac{1}{(k+1)^2}<\frac{1}{k(k+1)},$$ which is obvious.

Michael Rozenberg
  • 194,933
  • In your answer why did you use $ < $ and not $ \le $? –  Jun 29 '17 at 03:34
  • @Josh Mitkitzel We proved that in our statement should be $<$ for all $n\geq2$, but since for $n=1$ we get an equality, our statement is true with $\leq$ for all $n\geq1$. – Michael Rozenberg Jun 29 '17 at 03:43
  • So is it ok if I put $\le $? –  Jun 29 '17 at 03:46
  • @Josh Mitkitzel Yes, of course! If $\frac{1}{(k+1)^2}<\frac{1}{k(k+1)}$ then $\frac{1}{(k+1)^2}\leq\frac{1}{k(k+1)}$. But we can not change $\leq$ on $<$ in the given because it wold wrong. – Michael Rozenberg Jun 29 '17 at 03:48
2

You may use creative telescoping for proving a much tighter inequality.
If we set $H_n^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2}$, for any $n\geq 3$ we have: $$\begin{eqnarray*} H_n^{(2)} &=& \sum_{k=1}^{n}\frac{1}{k(k+1)}+\sum_{k=1}^{n}\frac{1}{k^2(k+1)}\\&=&\left(1-\frac{1}{n+1}\right)+\frac{1}{2}+\sum_{k=1}^{n-1}\frac{1}{(k+1)^2 (k+2)}\\&=&\frac{3}{2}-\frac{1}{n+1}+\sum_{k=1}^{n-1}\frac{1}{k(k+1)(k+2)}-\sum_{k=1}^{n-1}\frac{1}{k(k+1)^2(k+2)}\\&=&\frac{5}{3}-\frac{2n+1}{2n(n+1)}-\sum_{k=1}^{n-2}\frac{1}{(k+1)(k+2)^2(k+3)}\\ &\color{red}{\leq}&\color{red}{\frac{5}{3}-\frac{1}{n}} \end{eqnarray*}$$ and the given inequality for $n\in\{1,2\}$ can be easily checked by hand.
It is interesting to point out that this approach leads to a short proof of Stirling's inequality.

As an alternative, the convexity of $\frac{1}{x^2}$ over $\mathbb{R}^+$ leads, by the Hermite-Hadamard inequality, to:

$$ H_{n}^{(2)}=\frac{\pi^2}{6}-\sum_{m\geq n+1}\frac{1}{m^2}\color{red}{\leq} \frac{\pi^2}{6}-\int_{n+1}^{+\infty}\frac{dx}{x^2}=\frac{\pi^2}{6}-\frac{1}{n+1} $$ $$ H_{n}^{(2)}\color{blue}{\geq} \frac{\pi^2}{6}-\frac{1}{n+\frac{1}{2}}. $$

Jack D'Aurizio
  • 353,855