Science:Math Exam Resources/Courses/MATH103/April 2014/Question 08 (b)

MATH103 April 2014

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Question 08 (b)
Determine all values of $x$ such that the following series converges: $\sum _{n=0}^{\infty }{\frac {(n+1)(x+1)^{n}}{5^{n}(n+2)}}.$ Work must be shown for full marks. Simplify fully.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Use the ratio test.

Hint 2
Don’t forget to check the cases for which the ratio test is indeterminate by different means.

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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. Let $a_{n}={\frac {n+1}{5^{n}(n+2)}}$ be the coefficient of $(x+1)^{n}$ in the series. By the ratio test, the series converges for all $x$ such that $\lim _{n\to \infty }\left|{\frac {a_{n+1}(x+1)^{n+1}}{a_{n}(x+1)^{n}}}\right|<1$. Now ${\begin{aligned}{\frac {a_{n+1}(x+1)^{n+1}}{a_{n}(x+1)^{n}}}&={\frac {n+2}{5^{n+1}(n+3)}}\cdot {\frac {5^{n}(n+2)}{(n+1)}}(x+1)\\&={\frac {(n+2)^{2}}{5(n+3)(n+1)}}(x+1)\\&={\frac {n^{2}+4n+4}{5n^{2}+20n+15}}(x+1).\\&={\frac {(1/n^{2})(1+4/n+4/n^{2})}{(1/n^{2})(5+20/n+15/n^{2})}}(x+1)\end{aligned}}$ Thus, as n approaches infinity $\lim _{n\to \infty }\left|{\frac {a_{n+1}(x+1)^{n+1}}{a_{n}(x+1)^{n}}}\right|={\frac {1}{5}}|x+1|$ and so the series converges whenever ${\frac {1}{5}}|x+1|<1$. But ${\begin{aligned}{\frac {1}{5}}|x+1|<1&\Rightarrow -1<{\frac {1}{5}}(x+1)<1\\&\Rightarrow -5<x+1<5\\&\Rightarrow -6<x<4.\end{aligned}}$ Lastly, we check the boundary cases where $x=-6$ or $x=4$ (the ratio test is indeterminate for these cases). When $x=-6$, $a_{n}={\frac {(n+1)(-5)^{n}}{5^{n}(n+2)}}={\frac {n+1}{n+2}}(-1)^{n}.$ Thus, $\lim _{n\to \infty }a_{n}\neq 0$ (in fact, $a_{n}$ oscillates and does not converge), so the series diverges by the divergence test. When $x=4$, $a_{n}={\frac {(n+1)5^{n}}{5^{n}(n+2)}}={\frac {n+1}{n+2}},$ so $\lim _{n\to \infty }a_{n}=1\neq 0$ and the series diverges once again. We conclude that the only values for which the series converges are $-6<x<4$.

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