MATH101 April 2018
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Question 08 (ii)

Calculate $\displaystyle \sum _{n=2}^{\infty }\left({\frac {(3)^{n}}{2^{2n}}}+{\frac {1}{\sqrt {n+2}}}{\frac {1}{\sqrt {n+1}}}\right).$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Part of the series will telescope.

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Solution

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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 ${\textstyle {\color {blue}{\frac {9}{28}}{\frac {1}{\sqrt {3}}}}}$.

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