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This is a practice exercise from our class about proving inequalities using mathematical induction. I've been stuck on the last step for quite a while now.

This is the Question. "Prove that $\sum_{k=1}^n\frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}$ whenever $n$ is a positive integer."

This is my attempt to prove it.

Step 1: Base case $(n=1)$

$\sum_{k=1}^1\frac{1}{(1)^2}\le2-\frac{1}{1}$

$1\le1$

Step 2: Induction hypothesis

Assume that $\sum_{k=1}^n\frac{1}{k^2}\le 2-\frac{1}{n}$ is true for $n=c$

Then, $\sum_{k=1}^c\frac{1}{k^2}\le 2-\frac{1}{c}$

Step 3: prove that it is true for $n=c+1$

$\sum_{k=1}^{c+1}\frac{1}{k^2}$ $\le$ $2-\frac{1}{c}+\frac{1}{(c+1)^2}$

$\le 2+\frac{-(c+1)^2+c}{c(c+1)^2}$

$\le 2+\frac{-(c^2+2c+1)+c}{c(c+1)^2}$

$\le 2+\frac{-c^2-2c-1+c}{c(c+1)^2}$

$\le 2-\frac{c^2+c+1}{c(c+1)^2}$

$\le 2-\frac{c^2+c}{c(c+1)^2}-\frac{1}{c(c+1)^2}$

$\le 2-\frac{c(c+1)}{c(c+1)^2}-\frac{1}{c(c+1)^2}$

$\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$

I'm Stuck on this step.

Farshid Farhat
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Benny
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  • You want to prove some inequality involving $c$. When you write out the final solution you can make it look like a nice sequence of inequalities, but when you attempt to prove you just try to simplify it as possible until you see if you get something true you can go back from. – Michal Adamaszek Sep 16 '22 at 11:07
  • @trancelocation how come? – Benny Sep 16 '22 at 11:11
  • @MichalAdamaszek thank you for this, I will attempt the last step again – Benny Sep 16 '22 at 11:12
  • @trancelocation I overlooked it, thank you so much for this, I will edit my post now – Benny Sep 16 '22 at 11:35
  • Also: https://math.stackexchange.com/q/2488093/42969, https://math.stackexchange.com/q/638418/42969, https://math.stackexchange.com/q/4441933/42969. – Martin R Sep 16 '22 at 20:38

2 Answers2

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In step3, you added a new term $\frac{1}{c+1}$ (while $n=c+1$) to the both sides of the induction hypothesis ($\sum_{k=1}^{c}\frac{1}{k^2} \le 2-\frac{1}{c}$) and ended with:

$\sum_{k=1}^{c+1}\frac{1}{k^2} \le 2-\frac{1}{c}+\frac{1}{(c+1)^2} \le ... \le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$

Because $n=c+1$ and $n>1$, then $c>0$, then $\frac{1}{c(c+1)^2}>0$. Now we have:

$\sum_{k=1}^{c+1}\frac{1}{k^2} \le ... \le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2} \le 2-\frac{1}{c+1}$

And you are done with the deduction.

Farshid Farhat
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5

It is very straightforward induction. You just add

$\dfrac{1}{(n+1)^{2}}$ to both sides of $P(n)$ and you get

$2-\dfrac{1}{n}+\dfrac{1}{(n+1)^{2}}$ must be less than $2-\dfrac{1}{n+1}$ .

Making all the simplifications we get

$\dfrac{n+2}{(n+1)^{2}}<\dfrac{1}{n}$ which is obviously true. So we get

$P(n+1)$ and we are done!