Science:Math Exam Resources/Courses/MATH105/April 2013/Question 02 (b)

MATH105 April 2013

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Question 02 (b)
Find out whether the numerical series below converges or diverges. You should provide appropriate justification in order to receive credit. (b) $\sum _{k=1}^{\infty }{\frac {k^{10}10^{k}(k!)^{2}}{(2k)!}}\,.$

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
Given the appearance of factorial terms in the series, you might try applying the ratio 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. We apply the ratio test for this problem. Let $\quad c=\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|,\quad a_{k}={\frac {k^{10}10^{k}(k!)^{2}}{(2k)!}}.$ By the ratio test, if $c<1$, then the series converges. If $c>1$, the series diverges. If $c=1$, then no conclusion can be drawn. For $k>0$, all $a_{k}$ are positive so we can remove the absolute value signs in $c$ when evaluating it. ${\begin{aligned}c&=\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=\lim _{k\rightarrow \infty }{\frac {a_{k+1}}{a_{k}}}\\&=\lim _{k\rightarrow \infty }{\frac {(k+1)^{10}10^{k+1}((k+1)!)^{2}}{(2(k+1))!}}{\frac {(2k)!}{k^{10}10^{k}(k!)^{2}}}\\&=\lim _{k\rightarrow \infty }{\frac {(k+1)^{10}}{k^{10}}}{\frac {10^{k+1}}{10^{k}}}{\frac {((k+1)!)^{2}}{(k!)^{2}}}{\frac {(2k)!}{(2k+2)!}}\end{aligned}}$ Using the fact that $(k+1)!=(k+1)\cdot k!$ and $(2k+2)!=(2k+2)\cdot (2k+1)\cdot (2k)!$ we can simplify the fractions with factorial terms, giving us the following: ${\begin{aligned}c&=\lim _{k\rightarrow \infty }\left({\frac {k+1}{k}}\right)^{10}\cdot 10\cdot (k+1)^{2}\cdot {\frac {1}{(2k+2)(2k+1)}}\\&=\lim _{k\rightarrow \infty }\left(1+{\frac {1}{k}}\right)^{10}\cdot {\frac {10(k+1)^{2}}{(2k+2)(2k+1)}}\end{aligned}}$ Dividing both the numerator and denominator by $k^{2}$ gives: ${\begin{aligned}&=\lim _{k\rightarrow \infty }\left(1+{\frac {1}{k}}\right)^{10}{\frac {10(1+1/k)^{2}}{(2+2/k)(2+1/k)}}\\&={\frac {10}{4}}>1\end{aligned}}$ Since $c>1$, this series diverges by the ratio test.

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