Science:Math Exam Resources/Courses/MATH103/April 2015/Question 08

MATH103 April 2015

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Question 08
Find all $x$ such that the series $\sum _{n=0}^{\infty }{\frac {(n-1)^{3}(x-2)^{n}}{1+3^{n}}}$ converges?

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
Then, consider the divergence test.

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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. First, we use the ratio test. Since we have ${\begin{aligned}\lim _{n\to \infty }\left|{\frac {\frac {n^{3}(x-2)^{n+1}}{1+3^{n+1}}}{\frac {(n-1)^{3}(x-2)^{n}}{1+3^{n}}}}\right|&=\lim _{n\to \infty }\left|{\frac {n^{3}(x-2)^{n+1}}{1+3^{n+1}}}\right|\left|{\frac {1+3^{n}}{(n-1)^{3}(x-2)^{n}}}\right|=\lim _{n\to \infty }\left|{\frac {n^{3}}{(n-1)^{3}}}\right|\lim _{n\to \infty }\left|{\frac {1+3^{n}}{1+3^{n+1}}}\right||x-2|\\&=\lim _{n\to \infty }\left|{\frac {1}{(1-{\frac {1}{n^{3}}})^{3}}}\right|\lim _{n\to \infty }\left|{\frac {{\frac {1}{3^{n}}}+1}{{\frac {1}{3^{n}}}+3}}\right||x-2|=1\cdot {\frac {1}{3}}\cdot |x-2|={\frac {1}{3}}|x-2|,\end{aligned}}$ when ${\frac {1}{3}}|x-2|<1$ the series converges, while when ${\frac {1}{3}}|x-2|>1$ the series diverges. Since ${\frac {1}{3}}|x-2|<1$ is equivalent with $|x-2|<3$ and $-1<x<5$, we can rephrase that the series converges for $-1<x<5$ and diverges for $x<-1$ and $x>5$. Now we consider the endpoints; when $x=-1$. Plugging this into the summation we have $\sum _{n=0}^{\infty }{\frac {(n-1)^{3}(-1-2)^{n}}{1+3^{n}}}=\sum _{n=0}^{\infty }{\frac {(n-1)^{3}(-1)^{n}3^{n}}{1+3^{n}}}$. Then, we can observe that the sequence ${\frac {(n-1)^{3}(-1)^{n}3^{n}}{1+3^{n}}}$ doesn't converges to 0 as $n$ goes to infinity. Indeed, if we assume that it converges to 0, by limit rule we have the convergence of the sequence $(-1)^{n}(n-1)^{3}$ at infinity: $\lim _{n\to \infty }(-1)^{n}(n-1)^{3}=\lim _{n\to \infty }{\frac {\frac {(n-1)^{3}(-1)^{n}3^{n}}{1+3^{n}}}{\frac {3^{n}}{1+3^{n}}}}={\frac {\lim _{n\to \infty }{\frac {(n-1)^{3}(-1)^{n}3^{n}}{1+3^{n}}}}{\lim _{n\to \infty }{\frac {1}{{\frac {1}{3^{n}}}+1}}}}={\frac {0}{1}}=0.$ However, apparently, this is not true when we expand the sequence $a_{n}=(-1)^{n}(n-1)^{3}$: $a_{1}=0,\quad a_{2}=1,\quad a_{3}=-8,\quad a_{4}=27,\quad a_{5}=-64,\quad a_{6}=125,\cdots$ Therefore, by the divergence test, the given series at $x=-1$ diverges. On the other hand, the given series at $x=5$ also diverges. This follows from again the divergence test. Indeed, at $x=5$ the series can be written as $\sum _{n=0}^{\infty }{\frac {(n-1)^{3}3^{n}}{1+3^{n}}}$. Since the sequence ${\frac {(n-1)^{3}(-1)^{n}3^{n}}{1+3^{n}}}$ doesn't converges to 0 as $n$ goes to infinity, so is $\left|{\frac {(n-1)^{3}(-1)^{n}3^{n}}{1+3^{n}}}\right|={\frac {(n-1)^{3}3^{n}}{1+3^{n}}}$. Collecting all the information, the given series converges for $\color {blue}-1<x<5$.