How do I prove by induction that, for $n≥1, \sum_{r=1}^n \frac{1}{r(r+1)}=\frac{n}{n+1}$?

Hi can you help me solve this: I have proved that $p(1)$ is true and am now assuming that $p(k)$ is true. I just don't know how to show $p(k+1)$ for both sides?

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Do you mean $\sum_{r=1}^n \frac{1}{r(r+1)}=\frac{n}{n+1}$? –  Ben Millwood Jan 7 '13 at 21:45
Yes! sorry I do not know how to format like that. –  maxmitch Jan 7 '13 at 21:50
You can learn some here and at some of the references that you’ll find there. –  Brian M. Scott Jan 7 '13 at 21:52
Thanks I have just learnt it! :) –  maxmitch Jan 7 '13 at 22:03

Hint: $$\sum_{r=1}^{k+1}\frac{1}{r(r+1)}=\sum_{r=1}^{k}\frac{1}{r(r+1)}+\frac{1}{(k+1)((k+1)+1)}.$$ Now use the inductive hypothesis and see if that gets you anywhere.

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HINT:

$$\frac1{r(r+1)}=\frac1r-\frac1{r+1}$$

and

$$\frac{n}{n+1}=1-\frac1{n+1}\;.$$

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I suggest that the downvoter think a bit more about how these two observations can be incorporated into a proof by induction. (Indeed, one has only to look at André’s remark.) –  Brian M. Scott Jan 7 '13 at 22:35

If you want to do it by a conventional "blind" induction, suppose that for a certain $k$ we have $$\sum_1^k \frac{1}{r(r+1)}=\frac{k}{k+1}.\tag{1}$$ We want to prove that $$\sum_1^{k+1} \frac{1}{r(r+1)}=\frac{k+1}{k+2}.\tag{2}$$ Note that the left-hand side of $(2)$ is the left-hand side of $(1)$, plus $\dfrac{1}{(k+1)(k+2)}$.

So we want to prove that $$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}.\tag{3}$$ It seems reasonable to manipulate the left-hand side of $(3)$ and see whether we get the right-hand side. A common manipulation is to bring the expression to the common denominator $(k+1)(k+2)$. We get $$\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}.$$ This is equal to $\dfrac{k^2+2k+1}{(k+1)(k+2)}$.

But the numerator is equal to $(k+1)^2$. Cancel a $k+1$.

Remark: The algebra at the end is neater, and closer to the informal "telescoping" argument, if we observe that $\dfrac{1}{(k+1)(k+2)}=\dfrac{1}{k+1}-\dfrac{1}{k+2}$. Thus $$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+ \frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2}=\frac{k+1}{k+2}.$$

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$$\sum_{r=1}^{n}\frac{1}{r(r+1)}=\frac{n}{n+1}$$ for $n=1$ we have $\frac{1}{1(1+1)}=\frac{1}{1+1}$ suppose that $$\sum_{r=1}^{k}\frac{1}{r(r+1)}=\frac{k}{k+1}$$ then $$\sum_{r=1}^{k+1}\frac{1}{r(r+1)}=\sum_{r=1}^{k}\frac{1}{r(r+1)}+\frac{1}{(k+1)(k+2)}=$$ $$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=$$ $$=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}=\frac{(k+1)}{(k+1)+1}$$

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