How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ 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
$$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$
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$$ = \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$
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$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
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$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
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$$ = \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$
 A: [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.
A: Hint:
It is a telescoping sum since, for any $k>0$,
$$\frac 1{k(k+1)}=\frac 1k-\frac 1{k+1}.$$
A: 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
$$ \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)} = \frac{k^2+2k+1-k^2-2k}{(k+1)(k+1+1)}=?$$
A: Hint:

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

