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I' m doing this induction exercise: $\sum_{i=1}^{n}\left ( \frac{1}{i^{2}} +i\right )\leq \frac{n^{3}+n^{2}+4n-2}{2n}$ where $n\geq 1$

I ve proved step p(1) , now i m doing the induction step. I ve done few calculation for n+1: $\sum_{i=1}^{n}\left ( \frac{1}{i^{2}} +i\right )\leq \frac{n^{4}+3n^{3}+7n^{2}+7n}{2(n+1)^2}$

At this point i need help. I think yo subtract $ \frac{n^{3}+n^{2}+4n-2}{2n}$ to show the result is greater than 0

John Pi
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2 Answers2

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If you are allowed to use $\sum_{i=1}^ni=\frac{n(n+1)}2$ then, by Vishu's comment, the proposed duplicate will do.

If you are not then, after your (correct) calculations, there remains to check that $$\frac{n^{3}+n^{2}+4n-2}{2n}\le\frac{n^{4}+3n^{3}+7n^{2}+7n}{2(n+1)^2},$$ i.e. $$(n^{3}+n^{2}+4n-2)(n^2+2n+1)\le n(n^{4}+3n^{3}+7n^{2}+7n)$$ i.e. $$n^5+3n^4+7n^3+7n^2-2\le n^5+3n^4+7n^3+7n^2 $$ i.e. $$-2\le0,$$ which is true.

Anne Bauval
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    He did not claim he proved $A(n)\le\frac{n^{4}+3n^{3}+7n^{2}+7n}{2(n+1)^2}.$ He only computed $p(n+1)-\frac1{(n+1)^2}-(n+1)=\frac{n^{4}+3n^{3}+7n^{2}+7n}{2(n+1)^2},$ and concluded that in order to prove the induction step $A(n)\le p(n)\Rightarrow A(n+1)\le p(n+1)$, he only needed to prove that $p(n)\le \frac{n^{4}+3n^{3}+7n^{2}+7n}{2(n+1)^2},$ which I did. – Anne Bauval Jan 28 '23 at 18:34
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From Vishu's comment, it suffices to show that

$$ \sum_{i=1}^n{1\over i^2}\le2-\frac1n $$

This is trivial when $n=1$. Assume the inequality holds for $n$, so we have

\begin{aligned} \sum_{i=1}^{n+1}{1\over i^2} &\le2-\frac1n+{1\over(n+1)^2}=2-{(n+1)^2-n\over n(n+1)^2} \\ &=2-{n^2+n+1\over n(n+1)^2}<2-{n^2+n\over n(n+1)(n+1)}=2-{1\over n+1} \end{aligned}

Thus, the induction step is complete.

TravorLZH
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