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October 29th, 2017, 10:00 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 59 Thanks: 1  does this series converge or diverge:
sum of (2^(2n+1))/(2n+1)! from n=0 to infinity I have applied the ratio test to find that the limit=1 which tells me nothing about convergence or divergence, what else can i try to help find out more about this series? 
October 29th, 2017, 10:55 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,034 Thanks: 2342 Math Focus: Mainly analysis and algebra 
I claim that for sufficently large $k$, $$0 < \frac{2^k}{k!} < \frac{1}{k^2}$$ Would this help you if it were true? And if so, can you prove it? 
October 30th, 2017, 05:44 AM  #3 
Member Joined: Jan 2016 From: Blackpool Posts: 59 Thanks: 1 
thank you! I would solve this using the simple comparison test, since 1/k^2 is an example of a convergent series and is greater than 2^k/k! for all n>0 then 2^k/k! is also a convergent series, this helps a lot and I probably should have seen that the numerator and denominator are similar by letting k=(2n+1)

October 30th, 2017, 06:20 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,034 Thanks: 2342 Math Focus: Mainly analysis and algebra  
October 30th, 2017, 09:25 AM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,662 Thanks: 1327  Quote:
$\displaystyle \lim_{n \to \infty} \bigg \dfrac{2^{2(n+1)+1}}{[2(n+1)+1]!} \cdot \dfrac{(2n+1)!}{2^{2n+1}} \bigg$ $\displaystyle \lim_{n \to \infty} \bigg \dfrac{2^{2n+3}}{(2n+3)!} \cdot \dfrac{(2n+1)!}{2^{2n+1}} \bigg$ $\displaystyle \lim_{n \to \infty} \dfrac{4}{(2n+3)(2n+2)} = 0$  
October 30th, 2017, 09:53 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,034 Thanks: 2342 Math Focus: Mainly analysis and algebra 
\begin{align*} &&\frac{2^k}{k!} & = \frac{ k^2 2^k }{k!} \frac1{k^2} &= \frac{k}{k} \frac{k}{k1} \frac{2^k}{(k2)!} \frac1{k^2} \\ &&&= \frac{k}{2(k1)} \frac{2^{k+1}}{(k2)!} \frac1{k^2} &= 2^4 \frac{k}{2(k1)} \frac{2^{k3}}{(k2)!} \frac1{k^2} \\ &&&< 2^4 \frac11 \frac22 \frac23 \frac24 \frac25 \frac26 \frac27 \frac1{k^2} &\text{(for all $k > 8$)} \\ &&&= \frac11 \frac22 \frac23 \frac44 \frac45 \frac46 \frac47 \frac1{k^2} \\ &&&< \frac1{k^2} \\[8pt] &\text{so} & \frac{2^k}{k!} & < \frac1{k^2} &\text{(for all $k > 8$)} \end{align*} That's not a least lower bound, but it is a lower bound. 

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converge, diverge, sequence, series 
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