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I'm learning mathematical induction. I found this task, which I can't do properly:

Prove $\sum_{j=1}^{2n}\frac{1}{j(j+1)}=\frac{2n}{2n+1}$ for all $n\in\Bbb N$.

I know the rules $n = 1,\,n = k,\,n = k + 1$. I tried to do it when I put it in $n = 1,\,n = 2$, but it always comes to me differently $1/6 = 2/3$. Can someone explain to me how to do this :)

J.G.
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Jons Icey
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  • Please don't use pictures, use MathJax. Here is a tutorial. – Dietrich Burde May 08 '22 at 19:10
  • @DietrichBurde sorry, i will update :) – Jons Icey May 08 '22 at 19:11
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    The case $n=1$ is $\frac12+\frac16=\frac23$, which is correct; the case $n=2$ is $\frac12+\frac16+\frac{1}{12}+\frac{1}{20}=\frac45$, which is also correct. I'm not sure what went wrong when you tried checking early cases. – J.G. May 08 '22 at 19:16
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    As for how to do it, you need to prove (i) the $n=1$ case holds and (ii) if the $n=k$ case holds so does the $n=k+1$ case. The latter amounts to checking $\frac{2k}{2k+1}+\frac{1}{(2k+1)(2k+2)}+\frac{1}{(2k+2)(2k+3)}=\frac{2k+2}{2k+3}$. (You'll probably benefit from the observations $\frac{1}{j(j+1)}=\frac1j-\frac{1}{j+1},,\frac{2n}{2n+1}=1-\frac{1}{2n+1}$.) – J.G. May 08 '22 at 19:18
  • That looks like an answer. – marty cohen May 08 '22 at 19:31
  • By comparing your calculation for the $n=1$ case of $\frac 16$ to the correct calculation of $\frac 23$, as given by J.G., it looks like maybe you are misunderstanding either sigma notation, or exactly what the case for $n = 1$ is. – Paul Sinclair May 09 '22 at 13:18

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