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Prove with $n \ge 1$:

$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot4}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$

First, I prove it for $n=1$: $\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$ Which is true.

So I will now assume this: $\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$

And I want to prove it for $n+1$, i.e:

$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{((n+1)+1)2^{n+1}}$

This is how I tried to prove it:

$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} =$

$ 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$

Before continuing, I usually like to grab a calculator, give a value to $n$ and evaluate my current expression with the expression I want to reach. If both values are equal, it means I'm doing okay.

So I took $n=5$ and evaluated the expression I want to reach:

$1 - \frac{1}{(5+2)\cdot2^5+1} = \frac{447}{448}$ Then, still with $n = 5$ I evaluated my current expression: $1 - \frac{1}{(5+1)\cdot2^5}+\frac{5+3}{(5+1)(5+1)\cdot2^{5+1}} = \frac{575}{576}$

So I got $\frac{447}{448}$ for the expression I want to reach and $\frac{575}{576}$ for what I got so far. Something went wrong.

My problem with this is that I haven't done any calculations yet. All my steps so far were rather mechanical - things I always do with mathematical induction.

Maybe I simply evaluated them wrongly. But I can't see it - I've tested it many times already.

Why are both expressions different? They should be the same.

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    @AndréNicolas: Ah! You're right! Fixing now.2012-11-27

2 Answers 2

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Elevating comment to answer, at suggestion of OP:

In the last term of the last displayed equation, there is a $5+1$ where there should be a $5+2$. Jonas Meyer notes that this correction gets rid of the discrepancy.

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    Confirmed - it does work now. Thank you.2012-11-27
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There is a flaw in the logic in this posting:

begin quote:

First, I prove it for $n=1$: $\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$ Which is true.

end of quote

This is not valid reasoning. You're saying "Let's prove $A$, as follows: If $A$ then $3/4 = 3/4$, which is true." Very well then, I shall prove that Socrates was John F. Kennedy: If Socrates was John F. Kennedy, then Socrates died in 1963, and hence, Socrates was mortal. Which is true." To say "If B then A. And A is true. Therefore B." is a standard erroneous form of reasoning. Writers on logic since the time of Aristotle have identified this as a fallacy. A correct way of reasoning is this: $ \frac{1+2}{1(1+1)2^1} = \frac34 = 1 - \frac14 = 1-\frac{1}{(1+1)2^1}. $ Therefore the proposition in case $n=1$.

And then you repeat the same fallacy in your inductive step.

You should write "$=$" between things ALREADY KNOWN TO BE EQUAL. Then when you write $ A = B = C = \cdots = Z, $ then that proves that $A=Z$. You should not write $ A = Z $ therefore $ B=Y $ therefore $ C=X $ etc. etc. etc. etc.
Therefore $ 3=3 $ which is true. Therefore $A=Z$.

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    Ah! You're right about the $n=1$ case. But I'm a bit confused about the inductive step case. I mean, since I am *assuming* that $\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$ then I **do know** that $\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$, I guess.2012-11-27