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for $n\ge1$ let:$$P_n = (\frac21).(\frac54).(\frac{10}9).(\frac{17}{16})...(\frac{n^2+1}{n^2}) = \prod^n_{k=1}(\frac{k^2+1}{k^2}). $$ Prove by induction for $n\ge 2$: $$P_n \le 5 -\frac5n $$ I did ask a variation of this question before, the answers did help somewhat, but I still have trouble. So here is where I am stuck and don't know what to do:
Basis Step: Let $n = 2,$ then $$P_2 = \frac52 \ \le 5-\frac54 =\frac52$$ $$\frac52\le\frac52$$ is true.
Inductive Step: assume for $k\ge2$ $$P_k \le5-\frac5k$$ Need to prove that: $$P_{k+1} \le5-\frac{5}{k+1}$$Or $$P_{k+1} =P_k(\frac{(k+1)^2+1}{(k+1)^2})\le5-\frac{5}{k+1}$$So this is where I am stuck. I don't know how to prove it from here. The suggestion from here on didn't make sense to me (which was to replace $P_k$ with $5-\frac5k.)$ Any help would be appreciated.

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    $\begingroup$ Actually, $P_2=\frac 21 \cdot \frac 54 = \frac 52 \le \frac 52$. You lost the leading factor $\frac 21$ but it still works. $\endgroup$ – Ross Millikan Feb 25 '13 at 23:22
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You know that $P_k \le 5 -\frac{5}{k}$.

So $P_{k+1} =P_k(\frac{(k+1)^2+1}{(k+1)^2}) \le (5 -\frac{5}{k})(1+\frac1{(k+1)^2}) = 5(1-\frac1{k})(1+\frac1{(k+1)^2}) $ and you want this to be $\le 5(1-\frac1{k+1})$.

This is the same as ahowing $(1-\frac1{k})(1+\frac1{(k+1)^2}) \le 1-\frac1{k+1} $

or $1-\frac1{k}+\frac1{(k+1)^2}-\frac1{k(k+1)^2} \le1-\frac1{k+1} $

or $\frac1{(k+1)^2}-\frac1{k(k+1)^2} \le \frac1{k}-\frac1{k+1} $

or $\frac1{(k+1)^2}-\frac1{k(k+1)^2} \le \frac1{k(k+1)} $

which is clearly true since $\frac1{(k+1)^2} \le \frac1{k(k+1)} $.

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  • $\begingroup$ Thanks a lot for the step by step explanation. $\endgroup$ – user60334 Feb 26 '13 at 3:43
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To prove $$ P_k(\frac{(k+1)^2+1}{(k+1)^2})\le5-\frac{5}{k+1} $$ it is enough to prove $$ (5-\frac{5}{k})(\frac{(k+1)^2+1}{(k+1)^2})\le5-\frac{5}{k+1}. $$ Simplify this expression, and you will obtain what is required.

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