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How to prove if the following equality holds? $$\frac{k-1}{k}+\frac{1}{k(k+1)}=\frac{k}{k+1}$$

Maybe finding a common denominator would work, but I have no idea how to do it in this example.

I see that it holds for $k=1$: $$0+\frac12=\frac12=\frac1{1+1}.$$ It also holds for $k=2$: $$\frac12+\frac16 = \frac36+\frac16 = \frac46 = \frac23 = \frac2{2+1}$$
It also works for $k=3$: $$\frac23+\frac1{12}=\frac8{12}+\frac1{12}=\frac9{12}=\frac34=\frac3{3+1}$$

But I am not sure how to proceed if I am not working with numbers but with expressions.

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  • $\begingroup$ No magic. ${k-1\over k} = {k-1\over k} { k+1 \over k+1}$ and ${k \over k+1} = { k \over k+1} {k \over k}$. $\endgroup$
    – copper.hat
    Dec 23, 2014 at 20:27
  • $\begingroup$ you forgott $k\ne 0$ and $k \ne -1$ $\endgroup$ Dec 23, 2014 at 20:32
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    $\begingroup$ Note that the left-hand side is undefined if $k=0$, while the right-hand side is defined there. So equality does not quite hold. $\endgroup$ Dec 23, 2014 at 20:33

2 Answers 2

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Notice that $k(k+1)$ is a common multiple of $k$ and $k(k+1)$.

Hence, for $k\neq 0$ and $k \neq -1$, we have $$\frac{k-1}{k}+\frac{1}{k(k+1)}=\frac{k^2-1}{k(k+1)}+ \frac{1}{k(k+1)} = \frac{k^2}{k(k+1)} = \frac{k}{k+1}.$$

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Alex Silva
    Jan 21, 2015 at 22:07
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    $\begingroup$ I don't know Alex. The edit apparently does not satisfy everybody, and they still feel that this kind of a question is best left unanswered and removed. If you read meta for a while, you will learn that this issue is very contentious. There are good people on both sides, unfortunately also people unwilling to compromise at all on at least one side. We are trying to find a balance in the middle, but it is not very successful. You're welcome to join in the think-tank in meta looking for a resolution. Warning: certain people just want to argue, and are not really interested in improving the site. $\endgroup$ Jan 22, 2015 at 11:56
  • $\begingroup$ @Jyrki: "...on at least one side." That is interesting, in that I really can't figure out which side that would be. I wonder if you have an existence proof, or a particular side in mind. In any case, I hope it isn't quite true. $\endgroup$ Jan 23, 2015 at 14:20
  • $\begingroup$ @Jonas, I hope I'm wrong. I won't go into details now. May be later. $\endgroup$ Jan 23, 2015 at 18:11
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You're having trouble with finding the common denominator. Let's compare the two denominators.

$\frac{k-1}{k} + \frac{1}{k(k+1)}$ the difference between these two is one(on the right) is multiplied by $(k+1)$ and the other(the left) is not. To make the denominators common we multiply the numerator and the denominator of the left fraction by $\frac{k+1}{k+1}$ which if you think about it, is the same as multiplying it by one, which does not affect the value of the fraction. Doing this we get:

$\frac{k-1}{k} + \frac{1}{k(k+1)} = \frac{k^2-1}{k(k+1)} + \frac{1}{k(k+1)}$ $= \frac{k^2}{k(k+1)}$ (the $1$s cancel when we add them) $=\frac{k*k}{k(k+1)} = \frac{k}{k+1}$ (We cancel the $k$s on the top and bottom)

It is important that this equality holds almost everywhere. We cannot say it is valid at $k= 0,-1$ this is because these values of $k$ result in division by $0$ which we know is not allowed.

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