How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ [duplicate]

I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.

edit: edit: I am still stuck. Math isn't working out, I am making a mess with the constant edits, I will stop editing and not touch this so people can review the question. Sorry

a) Prove that P(1) is true
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$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$

Show that P(k+1) is true as well $$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$ $$~~$$ $$= \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$ $$~~$$ $$= \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$ $$~~$$ $$= \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$ $$~~$$ $$= \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$ $$~~$$ $$= \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$

marked as duplicate by Martin R, Key Flex, Martin Sleziak, Lord Shark the Unknown, max_zornFeb 4 at 7:08

• Try using ${1\over n(n+1)}=\frac1n-{1\over n+1}$ – saulspatz Feb 3 at 18:42
• I did and I ended up with $\frac{2}{k+1+1}$ instead of $\frac{1}{k+1+1}$. I can add my work for that too – Evan Kim Feb 3 at 18:43
• You'll have to show us that calculation before we can critique it. As for the proof you posted, I don't really understand what you are trying to do. Add some words, please. – saulspatz Feb 3 at 18:48
• – Martin R Feb 3 at 19:02
• – Martin R Feb 3 at 19:03

Hint: $$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{(k+1)(k+1)}{(k+1)(k+2)}.$$

Regarding your calculations, note that $$\frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)} = \frac{k^2+2k+1-k^2-2k}{(k+1)(k+1+1)}=?$$

• thank you, I see my problem now – Evan Kim Feb 4 at 0:53

[You can use the trick of observing that it's a telescoping sum, or you just blast ahead with a proof by induction. I think you're asking about the second approach, so here's an answer.]

It's like any other induction proof. You get to assume "P(k)", which means you get to assume $$\sum_{i=1}^{k} \frac{1}{i(i+1)} = \frac{k}{k+1}.$$ Now you want to show that "P(k+1)" is true: you want to show $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} \overset{?}{=} \frac{k+1}{k+2}.$$ Expand the left side: $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \sum_{i=1}^{k} \frac{1}{i(i+1)} + \frac{1}{(k+1)(k+2)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)},$$ where the last equality is by the inductive hypothesis. Now do some algebra: $$= \frac{k}{k+1} + \frac{1}{k+1} - \frac{1}{k+2} = 1 - \frac{1}{k+2} = \frac{k+1}{k+2}.$$ That completes the inductive step, and hence the proof.

Hint:

It is a telescoping sum since, for any $$k>0$$, $$\frac 1{k(k+1)}=\frac 1k-\frac 1{k+1}.$$

• I understand that part, but I think I am doing the math wrong in showing the equivalence when I am changing to greatest common denominators and combining the fractions. – Evan Kim Feb 3 at 18:54
• Precidely, since you obtain a telescoping sum, do not combine the fractions! – Bernard Feb 3 at 19:00

Hint:

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots+\frac{1}{n(n+1)}=\\\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)$$