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I have the equation: $$(1-\frac{1}{2^2})...(1-\frac{1}{n^2}) = \frac{n+1}{2n}$$ for n ≥ 2

Trying to prove by induction and I get the following equation.

$$\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2} = \frac{k+2}{2(k+2)}$$

I can't to simplify it to the final answer.

I multiplied $$\frac{k+1}{2k}.{(k+1)^2}$$ and $$\frac{k(k+2)}{(k+1)^2}.{2k}$$ but I know something is wrong somewhere!

BTW, this is part of the proof by Induction step!

misheekoh
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  • The original equation is not correct. The right hand side is always $\frac{1}{2}$ (unless $k=-2$). If you take $k=1$, for example, you get $\frac{7}{4}=\frac{1}{2}$. – Sam Weatherhog Nov 03 '15 at 22:01
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    @SamWeatherhog that indeed shows that the equation does not hold for $k=1$, it does not however make the whole equation "not correct". You cannot "disprove" an equation by finding a counter example. – null Nov 03 '15 at 22:07
  • @SamWeatherhog the equation on the RHS is correct. Yes I know that I'll get 1/2 for the RHS but is there no way that I can simplify the LHS to get RHS? If then, I guess RHS doesn't equal to LHS – misheekoh Nov 03 '15 at 22:08
  • @null yes the equation is only true for n≥2 – misheekoh Nov 03 '15 at 22:10
  • @null you can when there is no condition on the variable. At present, the question implies that the equation holds for all $k$ which it clearly does not. The equation doesn't hold for $k\ge2$ either. – Sam Weatherhog Nov 03 '15 at 22:13
  • @null the equation only holds for 3 values of $k$, two of which are complex. $k$ would need to satisfy $2k^3+5k^2+2k+1=0$ for the equation to hold. Since this is for a proof by induction, the equation is clearly wrong. – Sam Weatherhog Nov 03 '15 at 22:16
  • @misheekoh what is the actual induction question you are trying to prove? – Sam Weatherhog Nov 03 '15 at 22:17
  • @SamWeatherhog the information that this is part of a proof by induction has been edited into the question after I made my comment. Still, an equation in and of itself has no such thing as correctness, especially not depending on the desired result. – null Nov 03 '15 at 22:21
  • @SamWeatherhog just edited the question! – misheekoh Nov 03 '15 at 22:23

1 Answers1

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The statement is true for $n=2$. Suppose it holds for $n$; then $$ \left(1-\frac{1}{2^2}\right)\dots \left(1-\frac{1}{n^2}\right) \left(1-\frac{1}{(n+1)^2}\right)= \frac{n+1}{2n}\left(1-\frac{1}{(n+1)^2}\right) $$ The final term becomes $$ \frac{n+1}{2n}\frac{n(n+2)}{(n+1)^2}=\frac{(n+1)+1}{2(n+1)} $$ which is exactly what was to be proved.

(Note: I prefer avoiding the change from $n$ to $k$, do as you like better.)

You had a $+$ in $$ \frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2} $$ where multiplication should be used.

egreg
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