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.

  • $\begingroup$ 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 '14 at 20:27
  • $\begingroup$ you forgott $k\ne 0$ and $k \ne -1$ $\endgroup$ – Dr. Sonnhard Graubner Dec 23 '14 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. $\endgroup$ – André Nicolas Dec 23 '14 at 20:33

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 $$\frac{k-1}{k}+\frac{1}{k(k+1)}=\frac{k^2-1}{k(k+1)}+ \frac{1}{k(k+1)} = \frac{k^2}{k(k+1)} = \frac{k}{k+1}.$$

  • $\begingroup$ Why the downvote? $\endgroup$ – Alex Silva Jan 21 '15 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. $\endgroup$ – Jyrki Lahtonen Jan 22 '15 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. $\endgroup$ – Jonas Meyer Jan 23 '15 at 14:20
  • $\begingroup$ @Jonas, I hope I'm wrong. I won't go into details now. May be later. $\endgroup$ – Jyrki Lahtonen Jan 23 '15 at 18:11

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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