Given $a_1=1$ and $a_n=a_{n-1}+4$ where $n\geq2$ calculate, $$\lim_{n\to \infty }\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+\cdots+\frac{1}{a_na_{n-1}}$$
First I calculated few terms $a_1=1$, $a_2=5$, $a_3=9,a_4=13$ etc. So $$\lim_{n\to \infty }\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+\cdots+\frac{1}{a_na_{n-1}}=\lim_{n\to \infty }\frac{1}{5}+\frac{1}{5\times9}+\cdots+\frac{1}{a_na_{n-1}} $$
Now I got stuck. How to proceed further? Should I calculate the sum ? Please help.
HINT:
$$\dfrac4{a_ma_{m-1}}=\dfrac{a_m-a_{m-1}}{a_ma_{m-1}}=?$$
$a_m=1+4\cdot(m-1)=?$
Do you recognize the Telescoping series?
As lab bhattacharjee mentioned for every $n\in\mathbb{N}$, we have $$a_n-a_{n-1}=4$$ $$\begin{align} & {{I}_{n}}=\sum\limits_{i=1}^{n-1}{\frac{1}{{{a}_{i}}{{a}_{i+1}}}}=\sum\limits_{i=1}^{n-1}{\frac{1}{{{a}_{i+1}}-{{a}_{i}}}\left( \frac{1}{{{a}_{i}}}-\frac{1}{{{a}_{i+1}}} \right)}=\frac{1}{4}\sum\limits_{i=1}^{n-1}{\left( \frac{1}{{{a}_{i}}}-\frac{1}{{{a}_{i+1}}} \right)}=\frac{1}{4}\left( \frac{1}{{{a}_{i}}}-\frac{1}{{{a}_{n}}} \right) \\ & {{I}_{n}}=\frac{1}{4}\left( \frac{1}{{{a}_{1}}}-\frac{1}{{{a}_{1}}+4n-4} \right) \\ \end{align} $$ therefore $$\underset{n\to \infty }{\mathop{\lim }}\,{{I}_{n}}=\frac{1}{4{{a}_{1}}}=\frac{1}{4}$$
