Show that $\sum_{i=1}^{n}\frac{1}{i^2}\leq2-\frac{1}{n}$

Question

I am asked to prove by induction that $$\sum_{i=1}^{n}\frac{1}{i^2}\leq2-\frac{1}{n}$$

The base case is straightforward: $$\sum_{i=1}^{1}\frac{1}{1^2}\leq2-\frac{1}{1}$$

However, for the inductive step, the solution states that $n^2+2n \leq n^2 + 2n + 1$ or $(n+1)n + n \leq (n+1)^2$. This becomes $2 - \frac{1}{n} + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}$.

"From this, we can use the inductive hypothesis to see $$\sum_{i=1}^{n+1}\frac{1}{i^2}\leq + \frac{1}{(n+1)^2} \leq 2-\frac{1}{n} + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}"$$

Can you explain how we can come up with a quadratic equation on the right hand side and how that relates to the solution?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, May 03 '22 at 09:29
@Alejandro Welcome to Math SE. FYI, using Approach0, a question which basically has your procedure in reverse, i.e., going from $2 - \frac{1}{n} + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}$ to $n^2+2n \leq n^2 + 2n + 1$, is Is $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ correctly proved by induction?. ... — John Omielan, May 03 '22 at 10:01
@Alejandro (cont.) Other dup. questions are How does $\frac{-1}{k}+\frac{1}{(k+1)^2} \leq \frac{-1}{k+1}$ helps to show that $\sum\limits_{i=1}^k\frac1{i^2}\leq \ 2-\frac{1}{k}$, Standard Problem on Mathematical-Induction, Proof by induction involving inequalities and summation., Prove that for $ \displaystyle n≥1, \sum_ {i=1} ^{ n} \dfrac 1 {i^2} ≤ 2 - \dfrac 1 n$, Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$, Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ ... — John Omielan, May 03 '22 at 10:05
@Alejandro (cont.) and, finally, the first part of Prove summation inequality with induction. — John Omielan, May 03 '22 at 10:06

Show that $\sum_{i=1}^{n}\frac{1}{i^2}\leq2-\frac{1}{n}$

0 Answers0

Linked