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$$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$

Prove for $n=1$: $$\frac1{1\times2}=\frac1{1+1}=\frac12$$

Hip: $$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$

Demonstration: $$\frac1{n+1} + \frac1{(n+1)(n+2)}=\dots=\frac1{(n+1)+1}$$ My problem is that I can't find the correct algebra steps to get from the beginning of the demonstration to the end of the demonstration.

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  • $\begingroup$ Are you familiar with how induction works? If you are, it appears you have almost all of the work that would be needed. $\endgroup$ – Clayton Nov 13 '18 at 23:59
  • $\begingroup$ You may be struggling because the sum is $\frac{n}{n+1}$, not $\frac{1}{n+1}$. $\endgroup$ – Theo Bendit Nov 14 '18 at 0:26
  • $\begingroup$ Since the thing you are trying to prove is false (as Theo observed), you should ask yourself: what should you really be trying to prove. $\endgroup$ – GEdgar Nov 14 '18 at 0:27
  • $\begingroup$ Okay. Thanks everybody. So the problem might be the formula as Bendit said? $\endgroup$ – Sebas Martinez Santos Nov 14 '18 at 0:35
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Actually, this is incorrect. The correct answer should be $1-\frac1{n+1}$.

We can show that the above is true for $n=1$ easily. Now let us show $1-\frac1{n+1}+\frac1{(n+1)(n+2)}=1-\frac1{n+2}$

We can prove that $\frac1{n+1}-\frac1{n+2}=\frac {n+2}{(n+1)(n+2)}-\frac {n+1}{(n+1)(n+2)}=\frac{n+2-n-1}{(n+1)(n+2)}=\frac1{(n+1)(n+2)}$ and vice versa. The $\frac1{n+1}$ terms cancel out, giving us $1-\frac1{n+2}$

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$\frac {1}{1\times 2}=1-1/2$

$\frac{1}{2\times 3}=1/2 - 1/3$

........

$\frac {1}{n(n+1)} =1/n- \frac {1}{n+1}$

Add them up and cancel the middle terms to get $1-\frac {1}{n+1}=\frac {n}{n+1}$

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You can see immediately that what you have stated is false because the left side increases with increasing n while the right side decreases.

So, you have to find out what you really want to prove.

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