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Basically, the expression in a task looks like this:

I've made a thesis

Then I use an assumption, to change some part of thesis:

I replaced the red part with a right side of an assumption

Then I substracted parts, which was the same in both sides of the equation, and got this:

Finally equation

But this equation isn't true, it gives a 1 = 0 What should I change? Did I something wrong with thesis?

Elise Svart
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  • What does $\frac1{1\times2}+\frac1{1\times3}+\cdots+\frac1{(2n-1)(2n)}$ mean? If you put $n=1$ in $\frac1{(2n-1)(2n)}$, then you get indeed $\frac1{1\times2}$, but if you put $n=2$ there, then what you get is $\frac1{3\times4}$, not $\frac1{1\times3}$. – José Carlos Santos Oct 13 '19 at 09:16
  • Yeah, I made a mistake, now it's correct, thank you. – Elise Svart Oct 13 '19 at 09:24
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    My answer was an attempt to address the original version of the problem. The new one makes no sense either. Do you mean for the final term on the left to be $\frac{1}{n(2n-1)}$? – J.G. Oct 13 '19 at 09:28
  • Could you check this once again, please? – Elise Svart Oct 13 '19 at 09:45
  • @Tetris570 Who do you want to check what? My answer obtains the only way to write the right-hand side in the form $\sum_{k=1}^na_k$. – J.G. Oct 13 '19 at 10:50

1 Answers1

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Increasing $n$ from $k$ to $k+1$ adds $\frac{1}{(2k+1)(2k+2)}$ to the right-hand side, which is $0$ if $n=0$ as it's a sum of $n$ terms. Therefore, it's $\sum_{k=1}^n\frac{1}{2k(2k-1)}$, so the original problem has a misprint (the $\frac13$ should be $\frac{1}{12}$).

J.G.
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