$\lim_{n\to \infty}[\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n-1}a_{n}}].$

Question

Let $a_{1}=1$ and $a_{n}=a_{n-1}+4$ for $n\geq 2.$ I have to find $\lim_{n\to \infty}[\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n}a_{n-1}}].$ I tried it as

$\lim_{n\to \infty}[\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n}a_{n}}]=\lim_{n\to \infty}[\frac{1}{a_{1}[a_{1}+4]}+\frac{1}{a_{2}[a_{2}+4]}\cdot\cdot\cdot\frac{1}{a_{n-1}[a_{n-1}+4]}]=\frac{1}{4}\lim_{n\to\infty}[\sum \frac{1}{a_{n}}-\sum \frac{1}{a_{n}+4}].$

Now please help me to solve the problem. Thanks .

Answer 1

\begin{eqnarray}E(n)=\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n}a_{n-1}}& =&{1\over 4}\Big(\frac{a_2-a_1}{a_1a_2}+\frac{a_3-a_2}{a_{2}a_{3}}\cdot\cdot\cdot\frac{a_{n+1-a_n}}{a_{n}a_{n-1}}\Big)\\ & =&{1\over 4}\Big(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}\cdot\cdot\cdot\frac{1}{a_n}-\frac{1}{a_{n+1}}\Big)\\ & =&{1\over 4}\Big(\frac{1}{a_1}-\frac{1}{a_{n+1}}\Big)\\ \end{eqnarray}

since $a_n\to \infty$ we have $$\lim_{n\to \infty} E(n) = {1\over 4a_1} ={1\over 4}$$