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

Question

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.

Answer 1

What you really need is $2 − \frac{1}{k} + \frac{1}{(k+1)^2} \leq 2 − \frac{1}{(k+1)}$,

Answer 2

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 $\frac1{k(k-1)}$.)