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For every integer $n>1$, prove that :

$\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$

I don't seem to find any clue on how to relate the left side of the inequality to the right side.

I tried a little bit of AM-GM on the set {$\frac {1}{1^2}, \frac {1}{2^2},..,\frac {1}{n^2}$} :

$\sum_{k=1}^{n} {\frac {1} {k^2}} \geq n(\frac {1} {{(n!)}^2})^{1/n}$

Is that something helpful?

How to proceed ?

  • 0
    have you tried induction? – 2017-01-15
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    @Dr.SonnhardGraubner Not really... – 2017-01-15
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    For Q's like this, always try induction first. If it's insufficient then you have a much harder Q. – 2017-01-15

3 Answers 3

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for $n=2$ we have we have $$1+\frac{1}{4}>\frac{6}{5}$$ this is true since we have $$25>24$$ now we assume that is true: $$\sum_{k=1}^n\frac{1}{k^2}>\frac{3n}{2n+1}$$ and we have to prove that $$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3(n+1)}{2(n+1)+1}$$ we show that $$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3n}{2n+1}+\frac{1}{(n+1)^2}>\frac{3(n+1)}{2(n+1)+1}$$ this is true since $$\frac{3n}{2n+1}+\frac{1}{(n+1)^2}-\frac{3(n+1)}{2(n+1)+1}={\frac {n \left( 2+n \right) }{ \left( 2\,n+1 \right) \left( n+1 \right) ^{2} \left( 2\,n+3 \right) }} >0$$ since $n>1$

8

A cheeky solution:

$$\frac{3n}{2n+1}<\frac{3n}{2n}=\frac32$$

Likewise, for $n\ge7$,

$$\sum_{k=1}^n\frac1{k^2}\ge\sum_{k=1}^7\frac1{k^2}>\frac32>\frac{3n}{2n+1}$$

and one can numerically show this is true for $1

2

Induction, with the step being:

$$\sum_{k=1}^n\frac1{k^2}=\sum_{k=1}^{n-1}\frac1{k^2}+\frac1{n^2}\stackrel{Ind. \;Hyp.}>\frac{3(n-1)}{2(n-1)+1}+\frac1{n^2}=\frac{3n-3}{2n}+\frac1{n^2}=$$

$$=\frac{3n^2-3n+2}{2n^2}\ge\frac{3n}{2n+1}\iff6n^3-4n^2+n+2\ge6n^3$$

and the last inequality must be clear since

$$4n^2-n-2\ge0\;\;\; \forall\, n\ge\frac{1+\sqrt33}8\approx\frac{1+5.6}8\;,\;\;\text{and}\;\;0<\frac{1+5.6}8<1$$

so the inequality is true for all $\;n\in\Bbb N\;$ .

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    @SimpleArt Of course not...hehe. That inequality was before and I forgot to delete it. Thanks, editing. – 2017-01-15
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    Hm, you did the induction backwards. If it is true for some very large $n$, then it holds for all smaller $n\ge1$. Luckily, that's not hard, as seen from my answer. – 2017-01-15
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    @SimpleArt No, not really backwards: it is just that I assume truth for $\;n-1\;$ and prove for $\;n\;$ , whereas many people usually does assume for $\;n\;$ and proves for $\;n+1\;$ ... It is, of course, exactly the same after the first step of proving for $\;n=1\;$ . – 2017-01-15
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    Ah, ok. Flipping it like that flips my head XD – 2017-01-15