# How to solve this elementary induction proof: $\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\le\ 2-\frac{1}{n}$? [duplicate]

This is a seemingly simple induction question that has me confused about perhaps my understanding of how to apply induction

the question;

$$\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\ \le\ 2-\frac{1}{n},\ \forall\ n \ge1.$$

this true for $n=1$, so assume the expression is true for $n\le k$. which produces the expression,

$$\frac{1}{1^2} + \cdots + \frac{1}{k^2} \le\ 2-\frac{1}{k}.$$ now to show the expression is true for $k+1$,

$$\frac{1}{1^2}+\cdots+ \frac{1}{k^2} + \frac{1}{(k+1)^2} \le\ 2-\frac{1}{k}+\frac{1}{(k+1)^2}.$$

this the part I am troubled by, because after some mathemagical algebraic massaging, I should be able to equate,

$$2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\frac{1}{(k+1)},$$

which would prove the expression is true for $k+1$ and I'd be done. right? but these two are not equivalent for even $k=1$, because setting $k=1$ you wind up with $\frac{5}{4}=\frac{3}{2}$, so somewhere i am slipping up and I'm not sure how else to show this if someone has some insight into this induction that I'm not getting. thanks.

## marked as duplicate by Martin Sleziak, Najib Idrissi, drhab, Davide Giraudo, kingW3Feb 2 '15 at 10:20

• If you can't get $2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\frac{1}{(k+1)}$, the next natural thing is to try to get $2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq2-\frac{1}{(k+1)}.$ – Git Gud Jul 16 '14 at 21:22
• You need $$2 - \frac{1}{k} + \frac{1}{(k+1)^2} \leqslant 2 - \frac{1}{k+1},$$ not equality. – Daniel Fischer Jul 16 '14 at 21:22
• And the last equality is equivalent to $\frac1{(k+1)^2} \le \frac1k-\frac1{k+1}$. After using common denominator on the RHS, this inequality should be clear. – Martin Sleziak Jul 27 '14 at 6:19

What you really need is $2 − \frac{1}{k} + \frac{1}{(k+1)^2} \leq 2 − \frac{1}{(k+1)}$,
• ok so is it enough to say that the inequality $$2-\frac{1}{k}+\frac{1}{(k+1)^2} \le\ 2-\frac{1}{(k+1)}$$ is equivalent to the inequality $$\frac{1}{(k+1)}+\frac{1}{(k+1)^2} \le\ \frac{1}{k}$$, which is obvious enough to see? – user74091 Jul 16 '14 at 21:36
• Well, multiplying by $k(k+1)^2$, the inequality is seen to be equivalent to $k^2 + 2k \leq k^2 +2k +1$, which is now certainly obvius. – Marco Flores Jul 16 '14 at 21:46
Despite the fact that such solution is given in other posts (for example in this question about similar infinite series: Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$ it is used as an auxiliary result in some answers) it might be useful to mention the solution using telescoping sum in a question asking about finite sum.
Let us look at the sum starting with $k=2$: $$\sum\limits_{k=2}^n \frac1{k^2} \le \sum\limits_{k=2}^n \frac1{k(k-1)}.$$ After rewriting this as $\frac1{k(k-1)}=\frac1{k-1}-\frac1k$we see that on the RHS we get a telescoping sum, where many terms will cancel out: $$\sum_{k=2}^n \left(\frac1{k-1}-\frac1k\right)=(1-\frac12)+\left(\frac12-\frac13\right)+\dots+\left(\frac1{n-1}-\frac1n\right)=1-\frac1n$$ (We started with $k=2$ so that we do not get zero in the denominator in the expressions $\frac1{k-1}$ and $\frac1{k(k-1)}$.)