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I need to prove by induction that $Q_n \ge\frac{10}{3}-\frac5{3n}$ for $n\ge2$$$ Q_n = \left(\frac21\right)\cdot\left(\frac54\right)\cdot\left(\frac{10}{9}\right)\cdot\left(\frac{17}{16}\right)\cdot\ldots\cdot\left(\frac{n^2+1}{n^2}\right)=\prod^n_{k=1}\left(\frac{k^2+1}{k^2}\right)$$ I have been trying to come up with a formula for $Q_n$ , I think I need to know that before I can go ahead with the proof right? I did try writing $Q_n$ as: $$Q_n = \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\ldots\left(1+\frac{1}{n^2}\right)$$But that's not really helping me come up with a formula or the proof. Any help would be great.

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  • $\begingroup$ To prove $Q_n\geq x_n$, what you really need to do is to check it for $n=2$, and then show that $((n+1)^2+1)/(n+1)^2)x_n\geq x_{n+1}$, from which your induction step will follow. $\endgroup$ – Julien Feb 24 '13 at 1:43
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The point of induction is that you don't need to find a nicer formula for $Q_n$. You prove the base case for $n=2$ by plugging it in and verifying it. You then assume it's true for $Q_n$, and then you just need to show that $Q_{n+1}\geq \frac{10}{3}-\frac{5}{3(n+1)}$ follows from $Q_n\geq \frac{10}{3}-\frac{5}{3n}$. It does help to see that $Q_{n+1}=\left(1+\frac{1}{(n+1)^2}\right)Q_n$. Plug in the bound you have for $Q_n$ and then simplify. It amounts to showing that

$$\left(1+\frac{1}{(n+1)^2}\right)\left(\frac{10}{3}-\frac{5}{3n}\right)\geq\frac{10}{3}-\frac{5}{3(n+1)}$$

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  • $\begingroup$ In the last part why did you plug in $(\frac{10}{3}-\frac{5}{3n})$ for $Q_n$? $\endgroup$ – user60334 Feb 24 '13 at 2:13
  • $\begingroup$ @user60334: because $Q_n$ is bounded below by it (by assumption of induction) $\endgroup$ – Alex R. Feb 24 '13 at 2:57

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