Im having trouble with a proof by induction

Question

the question is:

let $$a_{n} = 1$$ and $$a_{n+1}=\frac{a_{n}}{1+(n+1)a_{n}}$$ for each natural number n. prove by induction that $$a_{n} = \frac{2}{n(n+1)}$$ for every natural number. and deduce that $$a_{n}\rightarrow 0.$$I have shown that when n = 1 the theorem is true, and assumed that it is also true for n = k
and when i try prove the case for n=k+1 I get: $$a_{k+1}=\frac{\frac{2}{k(k+1)}}{1+(k+1)\frac{2}{k(k+1)}}$$
its at this point that I'm getting stuck.
do i need to try and manipulate the equation to get:$$a_{k+1}=\frac{2}{(k+1)(k+2)}?$$
Id appreciate any help, cheers guys.

Answer 1

You're almost there; you've done nearly everything. Your expression is

$$a_{k+1} = \frac{\frac{2}{k(k+1)}}{1+(k+1)\frac{2}{k(k+1)}} = \frac{\frac{2}{k(k+1)}}{1+\frac{2}{k}} = \frac{2}{k(k+1)+2(k+1)} = \frac{2}{(k+2)(k+1)}$$

as desired.

Answer 2

Write $b_n = \dfrac{1}{a_n}$. Then $$ \frac{1}{b_{n+1}} = \frac{\frac{1}{b_{n}}}{1+\frac{n+1}{b_{n}}} = \frac{1}{b_n+n+1} $$ and so $$ b_{n+1} = b_n+n+1 $$ Thus, $$ b_n = \sum_{k=1}^n k = \frac{n(n+1)}{2} $$