Science:Math Exam Resources/Courses/MATH103/April 2017/Question 01 (b)

MATH103 April 2017

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[hide] Question 01 (b)
Determine whether the following series converge or diverge. Check appropriate box. converging diverging (i) $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ (ii) $\sum _{n=1}^{\infty }{\frac {-n^{2}}{1+n^{2}}}$ (iii) $\sum _{n=1}^{\infty }{\frac {n^{2}}{n!}}$ (iv) $\sum _{n=0}^{\infty }{\frac {1}{\sqrt {3n+4}}}$

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[show] Hint
Remember some critical rules we use in series. The divergence test, ratio test, $p$-series test, and limit comparison test are useful for this problem.

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[show] Solution

(i) from p-test, p>1 then it is converge. (ii) We use the divergence test. First, compute $\lim _{n\to \infty }{\frac {-n^{2}}{1+n^{2}}}$ . The limit can be written as $\lim _{n\to \infty }{\frac {-1}{1/n^{2}+1}}$ We have $\lim _{n\to \infty }({\frac {1}{n}})^{2}=0$, and therefore $\lim _{n\to \infty }{\frac {-1}{1/n^{2}+1}}=-1$ This implies that by the sequence is divergent by the divergence test. (iii) We use the ratio test. The ratio of the n+1 term to th n term is ${\frac {\frac {(n+1)^{2}}{(n+1)!}}{\frac {n!}{n^{2}}}}$ we therefore need to compute $\lim _{n\to \infty }{\frac {(n+1)^{2}n!}{n^{2}(n+1)!}}.$ This simplifies to $\lim _{n\to \infty }{\frac {(n+1)^{2}}{n^{2}}}\cdot {\frac {1}{n}}.$ The limit of ${\frac {(n+1)^{2}}{n^{2}}}$ is 1, and the limit of 1/n is zero, $\lim _{n\to \infty }{\frac {(n+1)^{2}}{n^{2}}}\cdot {\frac {1}{n}}=0$ and the series converges by the ratio test. (iv) We will use the limit comparison test here. Letting $a_{n}$ be the sequence ${\frac {1}{\sqrt {3n+4}}}$ and letting $b_{n}$ be the series ${\frac {1}{\sqrt {n}}},$ we have ${\begin{aligned}\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}&=\lim _{n\to \infty }{\frac {\sqrt {n}}{\sqrt {3n+4}}}\\&={\frac {1}{\sqrt {3+{\frac {4}{n}}}}}&={\frac {1}{\sqrt {3}}}\end{aligned}}$ this is finite and nonzero, so the limit comparison test says that $\sum _{n=1}^{\infty }a_{n}$ and $\sum _{n=1}^{\infty }b_{n}$ both converge or both diverge. But $\sum _{n=1}^{\infty }b_{n}$ diverges by the $p$-series test, so $\sum _{n=1}^{\infty }a_{n}$ does too. $\color {blue}Answer:$' converging diverging (i) $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ yes no (ii) $\sum _{n=1}^{\infty }{\frac {-n^{2}}{1+n^{2}}}$ no yes (iii) $\sum _{n=1}^{\infty }{\frac {n^{2}}{n!}}$ yes no (iv) $\sum _{n=0}^{\infty }{\frac {1}{\sqrt {3n+4}}}$ no yes