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Prove using induction that

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{n^2} \le 2-\frac{1}{n}$$

for all positive whole numbers $n$.

I began by showing that it is true for $n=1$

I then assumed that it is true for $n=p$ $$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} = \sum_{k=1}^p \frac{1}{k^2} \le 2-\frac{1}{p}$$

I now want to show that it is true for $n=p+1$

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2}= \sum_{k=1}^{p+1} \frac{1}{k^2} $$

If I add $\frac{1}{(p+1)^2}$ to $\sum_{k=1}^{p} \frac{1}{k^2}$, I will then get

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2} \le 2-\frac{1}{p} + \frac{1}{(p+1)^2}$$

If this is true then $$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$ or $$-\frac{1}{(p+1)}=-\frac{1}{p} + \frac{1}{(p+1)^2}$$ $$0= \frac{1}{(p+1)} -\frac{1}{p} + \frac{1}{(p+1)^2}$$ $$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$ $$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$ $$0= \frac{p(p+1)-(p+1)^2 + p}{p(p+1)^2}$$ $$0= \frac{(p^2 + p) - (p^2 + 2p + 1) + p}{p(p+1)^2}$$ $$0= \frac{-1}{p(p+1)^2}$$ This is invalid. I am not sure where I have made a mistake but I think it is $$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$ It then must be that $$-\frac{1}{(p+1)} < -\frac{1}{p} + \frac{1}{(p+1)^2}$$ $$0 \le \frac{1}{(p+1)} - \frac{1}{p} + \frac{1}{(p+1)^2}$$ $$ 0< \frac{-1}{p(p+1)^2}$$ which is true for all positive whole numbers $p$. I am pretty sure it is proved now but I would be happy if someone can confirm this.

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marked as duplicate by Martin R, Cyclohexanol., Dave, Leucippus, Namaste Oct 25 '17 at 0:53

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  • $\begingroup$ No need for equality as a special case. Just use $\le$ instead of $<$. $\endgroup$ – quasi Oct 24 '17 at 19:26
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    $\begingroup$ How could $\frac{-1}{p(p+1)^2}$ be a positive? The last line $\endgroup$ – Ameryr Oct 24 '17 at 19:27
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    $\begingroup$ You wrote: "If this is true then $$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}\text{"}$$ But what you really need is $$2-\frac{1}{p} + \frac{1}{(p+1)^2} \leq 2-\frac{1}{(p+1)}$$ $\endgroup$ – Bungo Oct 24 '17 at 19:29
  • $\begingroup$ @Bungo You can write $\frac{1}{p+1}$. You don't need the parentheses $\frac{1}{(p+1)}$. $\endgroup$ – user236182 Oct 24 '17 at 19:35
  • $\begingroup$ @Ameryr Ah I see the mistake now $\endgroup$ – E.Bob Oct 24 '17 at 19:38
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I think the following induction a bit of better. $$\sum_{k=1}^n\frac{1}{k^2}=1+\sum_{k=2}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1+1-\frac{1}{n}=2-\frac{1}{n}.$$

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  • $\begingroup$ Pure (math) poetry! $\endgroup$ – Pieter21 Oct 24 '17 at 19:42
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    $\begingroup$ It's a nice proof, but where's the induction? $\endgroup$ – Bungo Oct 24 '17 at 20:04
  • $\begingroup$ @Bungo We can understand what is it $\sum\limits_{k=1}^n(...)$ by induction. $\endgroup$ – Michael Rozenberg Oct 24 '17 at 20:06
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It can be much simpler.

You have equality for $n = 1$.

Adding a new $n+1$ term adds $\frac{1}{(n+1)^2}$ to LHS and $\frac{1}{n(n+1)}$ to the RHS. Guess which one is smaller for all $n>1$?!

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  • $\begingroup$ That is a lot easier. Thanks! $\endgroup$ – E.Bob Oct 24 '17 at 19:39

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