I was just studying the 'Method of Mathematical Induction' by I. S. Sominskii when this example came on: $$(\dots)$$
$$S_k=\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\dots+\frac{1}{k(k+1)}=\frac{k}{k+1}$$
$$(\dots)$$
So my problem is that I can't understand how to get to that result. Probably partial fractions???
Could someone please, explain me.
Thank you very much!
On
We have the definition: $\displaystyle\forall k\in\Bbb N^+~.~S_k:=\sum_{n=1}^k \dfrac{1}{n\,(n+1)}$
$$S_k=\dfrac 1{(1)(2)}+\dfrac{1}{(2)(3)}+\cdots+\dfrac{1}{(k)(k+1)}$$
We have a hypothesis that: $\forall k\in\Bbb N^+~.~ S_k=\dfrac{k}{k+1}$
Proof by mathematical induction:
If the hypothesis holding for any positive natural number $k$ does imply that it holds for $k+1$, and the hypothesis does hold when $k=1$ , then the hypothesis will hold for any positive natural number.
$$\begin{split}&S_1=\dfrac 1{1+1}&&\text{by the definition}\\&\forall k\in\Bbb N^+~.\left[ S_{k}=\dfrac{k}{k+1}\to S_{k+1}=\dfrac{k+1}{(k+1)+1}\right]&\qquad&\bigstar\text{ show this}\\\hline\therefore~ &\forall k\in\Bbb N^+~.~ S_k=\dfrac{k}{k+1}\end{split}$$
Well, by definition $S_{k+1}=S_k+\dfrac{1}{(k+1)\,(k+2)}$ . So you just need to show $$\dfrac{k}{k+1}+\dfrac{1}{(k+1)\,(k+2)}=\dfrac{k+1}{k+2}$$
Algebra kadabra!
Using from comment $$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$$ we have $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{k(k+1)}=\\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{k}-\frac{1}{k+1}= \\ =1-\frac{1}{k+1}=\frac{k}{k+1}$$