# Science:Math Exam Resources/Courses/MATH101/April 2018/Question 08 (ii)/Solution 1

We start by splitting the series

$\sum _{n=2}^{\infty }\left({\frac {(-3)^{n}}{2^{2n}}}+{\frac {1}{\sqrt {n+2}}}-{\frac {1}{\sqrt {n+1}}}\right)=\sum _{n=2}^{\infty }{\frac {(-3)^{n}}{2^{2n}}}+\sum _{n=2}^{\infty }\left({\frac {1}{\sqrt {n+2}}}-{\frac {1}{\sqrt {n+1}}}\right).$ Now the second series will telescope. Observe that for any positive integer ${\textstyle N}$ , we have
$\sum _{n=2}^{N}\left({\frac {1}{\sqrt {n+2}}}-{\frac {1}{\sqrt {n+1}}}\right)={\bigg (}{\frac {1}{\sqrt {4}}}-{\frac {1}{\sqrt {3}}}{\bigg )}+{\bigg (}{\frac {1}{\sqrt {5}}}-{\frac {1}{\sqrt {4}}}{\bigg )}+\cdots +{\bigg (}{\frac {1}{\sqrt {N+2}}}-{\frac {1}{\sqrt {N+1}}}{\bigg )}=-{\frac {1}{\sqrt {3}}}+{\frac {1}{\sqrt {N+2}}},$ so we have
$\lim _{N\to \infty }\sum _{n=2}^{N}\left({\frac {1}{\sqrt {n+2}}}-{\frac {1}{\sqrt {n+1}}}\right)=\lim _{N\to \infty }{\bigg [}-{\frac {1}{\sqrt {3}}}+{\frac {1}{\sqrt {N+2}}}{\bigg ]}=-{\frac {1}{\sqrt {3}}}.$ It remains to calculate the series

$\sum _{n=2}^{\infty }{\frac {(-3)^{n}}{2^{2n}}}=\sum _{n=2}^{\infty }{\frac {(-3)^{n}}{(2^{2})^{n}}}=\sum _{n=2}^{\infty }{\bigg (}{\frac {-3}{4}}{\bigg )}^{n}=-1+{\frac {3}{4}}+\sum _{n=0}^{\infty }{\bigg (}{\frac {-3}{4}}{\bigg )}^{n}=-{\frac {1}{4}}+\sum _{n=0}^{\infty }{\bigg (}{\frac {-3}{4}}{\bigg )}^{n}.$ Now the series is a geometric series, so using the formula for the geometric series we get that
$\sum _{n=0}^{\infty }{\bigg (}{\frac {-3}{4}}{\bigg )}^{n}={\frac {1}{1+{\frac {3}{4}}}}={\frac {4}{7}}.$ It follows that
$\sum _{n=2}^{\infty }\left({\frac {(-3)^{n}}{2^{2n}}}+{\frac {1}{\sqrt {n+2}}}-{\frac {1}{\sqrt {n+1}}}\right)={\frac {4}{7}}-{\frac {1}{4}}-{\frac {1}{\sqrt {3}}}={\frac {9}{28}}-{\frac {1}{\sqrt {3}}}.$ Answer: The correct answer is ${\textstyle {\color {blue}{\frac {9}{28}}-{\frac {1}{\sqrt {3}}}}}$ .