Can I show $\sum_{i=1}^n 1/{i^2} \le 2$ without calculus?

Question

I'm solving a number theory problem and it suffices to show that $$\sum_{i=1}^n \frac {1}{i^2} \le 2$$ In fact, I only need $$\sum_{d|n} \frac{1}{d^2} \le 2$$ Trying wolfram alpha suggests they are true but I don't know how to prove. A proof without calculus(which I think may be used) is preferable since I haven't learned it yet. Thank you for all helps:)

Answer 1

Since $x \mapsto \dfrac1{x^2}$ is a monotone decreasing function over $[1,\infty)$, then we have $$ \sum_{n=1}^N\frac1{n^2}\leq1+\int_1^N\frac1{x^2}dx=1+\left[-\frac1x \right]_1^N=2-\frac1N, \quad N\geq1. $$ From which you deduce easily that

$$ \sum_{d|N} \frac1{d^2}\leq\sum_{n=1}^N\frac1{n^2}\leq 2 $$

as announced.

Alternatively, you may use a telescoping sum: $$ \frac1{n^2}\leq \frac1{n(n-1)}=\frac1{n-1}-\frac1n,\qquad n\geq2, $$ giving $$ \sum_{n=1}^N\frac1{n^2}\leq1+\sum_{n=2}^N\left(\frac1{n-1}-\frac1n \right)=1+1-\frac1N\leq 2. $$

Answer 2

For $2^i\le j< 2^{i+1}$, we have $\frac{1}{j^2}\le\frac{1}{2^{2i}}$. As such, if we take $N=2^k-1$ for some $k$, then $$\sum\limits_{n=1}^{N}{\frac{1}{n^2}} = \sum\limits_{i=1}^{k-1}{\sum\limits_{j = 2^i}^{2^{i+1}-1}{\frac{1}{j^2}}}\le\sum\limits_{i=1}^{k-1}{\sum\limits_{j=2^i}^{2^{i+1}-1}{\frac{1}{2^{2i}}}} = \sum\limits_{i=1}^{k-1}{\frac{2^i}{2^{2i}}}=\sum\limits_{i=1}^{k-1}{\frac{1}{2^i}} = 2 - \frac{1}{2^{k-1}}\le 2. $$ Since the sum is monotonically increasing, the result follows for all $N$ (by looking at the next highest power of 2, minus 1).

This logic is quite similar to a common proof showing the divergence of the harmonic series.

Answer 3

By Taylor Formula for function $sinx$ i have $$sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-....$$ The equation $$sin x = 0$$ Have all roots are $k.\pi$ so we can delete $0$ ( in fact because it's a polynomial) $$1-\frac{x^2}{3!}+\frac{x^4}{5!}-...=\prod ( 1 - \frac{x^2}{k^2.\pi^2})$$ Compare coefficient of $x^2$ then $$\sum \frac{1}{n^2}= \frac{\pi^2}{6}<2$$