MATH105 April 2012
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Question 02 (c)

This problem contains three numerical series. For each of them, ﬁnd out whether it converges or diverges. You should provide appropriate justiﬁcation in order to receive credit.
 $\sum _{k=1}^{\infty }{\frac {2^{k}(k!)^{2}}{(2k)!}}$
Recall that $k!=1\cdot 2\cdot 3\cdot \ldots \cdot k$.

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

Whenever factorials are involved you should try to the ratio test.

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Solution

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Series of this form suggest the ratio test: given powers such as 2^{k} and factorials, we know that in taking the ratio of successive terms, many simplifications are possible.
Let us denote $a_{k}={\frac {2^{k}(k!)^{2}}{(2k)!}}$ so that $a_{k+1}={\frac {2^{k+1}((k+1)!)^{2}}{(2(k+1))!}}$.
The ratio test requires us to compute the limit
 $L=\lim _{k\rightarrow \infty }{\frac {a_{k+1}}{a_{k}}}$
Theory tells us that
 if $0\leq L<1$ then the series converges;
 if $\displaystyle L=1$, we cannot conclude anything from the test;
 if $\displaystyle L>1$ then the series diverges.
We need to be careful with the factorials (!). A useful thing to recall is that $(k+1)!=(k+1)k!$.
We will compute:
 ${\begin{aligned}L&=\lim _{k\rightarrow \infty }\left{\frac {a_{k+1}}{a_{k}}}\right=\lim _{k\rightarrow \infty }\left{\frac {\frac {2^{k+1}((k+1)!)^{2}}{(2(k+1))!}}{\frac {2^{k}(k!)^{2}}{(2k)!}}}\right\\&=\lim _{k\rightarrow \infty }{\frac {2^{k+1}((k+1)!)^{2}(2k)!}{2^{k}(k!)^{2}(2(k+1))!}}\\&=\lim _{k\rightarrow \infty }{\frac {2^{k+1k}\times [(k+1)k!]^{2}\times (2k)!}{(k!)^{2}\times (2k+2)!}}\\&=\lim _{k\rightarrow \infty }{\frac {2(k+1)^{2}(2k)!}{(2k+2)!}}\\&=\lim _{k\rightarrow \infty }{\frac {2(k+1)^{2}(2k)!}{(2k+2)(2k+1)(2k)!}}\\&=2\lim _{k\rightarrow \infty }{\frac {(k+1)^{2}}{(2k+2)(2k+1)}}\\&=2\lim _{k\rightarrow \infty }{\frac {k^{2}+2k+1}{4k^{2}+6k+2}}\\&=2\lim _{k\rightarrow \infty }{\frac {k^{2}/k^{2}+2k/k^{2}+1/k^{2}}{4k^{2}/k^{2}+6k/k^{2}+2/k^{2}}}\\&=2({\frac {1}{4}})={\frac {1}{2}}<1\end{aligned}}$
As $0\leq L<1$, we determine the series converges.

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