proof by induction (summations)

Question

I'm trying to use induction to prove that

$$\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+...+\frac{1}{(n)(n+1)}=1-\frac{1}{n+1}.$$

This is what I have so far:

Base Case: $n=1$. We get $$\sum_{i=1}^{k} \frac{1}{(i)(i+1)} = 1-\frac{1}{(k+1)} \implies \frac{1}{2} = \frac{1}{2}.$$

End Goal (what we want to show): $$\sum_{i=1}^{k+1} \frac{1}{(i)(i+1)} = 1-\frac{1}{k+2}.$$

So,

$$\begin{align} &\sum_{i=1}^{k} \frac{1}{(i)(i+1)} + \frac{1}{(k+1)(k+2)}\\ &=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}\\ &= 1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\ &= 1-\frac{(k+2)+1}{(k+1)(k+2)} \end{align}$$

But I don't know where to go from here. I can't cancel the $k+2$'s because I need one in the denominator so the LHS = RHS, but if I expand the numerator to $(k+1)+2$ and then cancel the $k+1$'s I'm left with a $2$ in the numerator when I need a $1$. Any suggestions?

Answer 1

You are almost there. The trick goes as by writing $$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2}$$ so that $$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}=1-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2} $$

Answer 2

You're mistake is in going from the second to last line to the last line. You subtracted both terms when the last term should have been added. The numerator in the final line should be $k+1$ instead of $k+3$. This will give you the desired result.

Answer 3

You messed up the fiddly diddly. The last line should be:$$=1-\frac{(k+2)-1}{(k+1)(k+2)}$$Should be OK now.