Science:Math Exam Resources/Courses/MATH101/April 2015/Question 11 (a)

MATH101 April 2015

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Question 11 (a)
(a) Find the value of the convergent series $\sum _{n=2}^{\infty }\left({\frac {2^{n+1}}{3^{n}}}+{\frac {1}{2n-1}}-{\frac {1}{2n+1}}\right)$ (simplify your answer completely).

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
Split the series into the sum of a geometric series and a telescoping series.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We rewrite the series as follow $\sum _{n=2}^{\infty }\left({\frac {2^{n+1}}{3^{n}}}+{\frac {1}{2n-1}}-{\frac {1}{2n+1}}\right)=\sum _{n=2}^{\infty }{\frac {2^{n+1}}{3^{n}}}+\sum _{n=2}^{\infty }{\frac {1}{2n-1}}-{\frac {1}{2n+1}}$ Now we compute each of these series: The first series is a geometric series, so we use the formula for the limit of a geometric series which is $\sum _{n=0}^{\infty }a\,r^{n}={\frac {a}{1-r}}$ for $|r|<1$, but first we simplify the series to write it in the suitable form to find the first term and the ratio: $\sum _{n=2}^{\infty }{\frac {2^{n+1}}{3^{n}}}=2\sum _{n=2}^{\infty }{\frac {2^{n}}{3^{n}}}=2\left({\frac {4}{9}}+{\frac {8}{27}}+...\right)=2\sum _{n=0}^{\infty }{\frac {4}{9}}\left({\frac {2}{3}}\right)^{n}=2\ {\frac {\frac {4}{9}}{1-{\frac {2}{3}}}}=2{\frac {\frac {4}{9}}{\frac {1}{3}}}={\frac {8}{3}}$ ${\begin{aligned}\sum _{n=2}^{\infty }{\frac {1}{2n-1}}-{\frac {1}{2n+1}}&=\left({\frac {1}{3}}-{\frac {1}{5}}\right)+\left({\frac {1}{5}}-{\frac {1}{7}}\right)+\left({\frac {1}{7}}-{\frac {1}{9}}\right)+...+\left({\frac {1}{2n-3}}-{\frac {1}{2n-1}}\right)+\left({\frac {1}{2n-1}}-{\frac {1}{2n+1}}\right)\\&={\frac {1}{3}}-{\frac {1}{2n+1}}\longrightarrow {\frac {1}{3}}\qquad \qquad {\text{when}}\ n\rightarrow \infty \end{aligned}}$ $\sum _{n=2}^{\infty }\left({\frac {2^{n+1}}{3^{n}}}+{\frac {1}{2n-1}}-{\frac {1}{2n+1}}\right)={\frac {8}{3}}+{\frac {1}{3}}={\frac {9}{3}}=\color {blue}3$