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

Question

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
$~$
$~$
$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$

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

Answer 1

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

Answer 2

Hint:

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

Answer 3

Hint:

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

Answer 4

[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.