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 October 29th, 2017, 09:00 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 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, 09:55 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,658 Thanks: 2635 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? Thanks from Jaket1
 October 30th, 2017, 04:44 AM #3 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 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, 05:20 AM   #4
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Quote:
 Originally Posted by Jaket1 ... and is greater than 2^k/k! for all k>0
Not all $k > 0$. It's not true for $k=1$ for example.

October 30th, 2017, 08:25 AM   #5
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Quote:
 Originally Posted by Jaket1 sum of (2^(2n+1))/(2n+1)! from n=0 to infinity I have applied the ratio test to find that the limit=1 ...
you may want to recheck ...

$\displaystyle \lim_{n \to \infty} \bigg| \dfrac{2^{2(n+1)+1}}{[2(n+1)+1]!} \cdot \dfrac{(2n+1)!}{2^{2n+1}} \bigg|$

$\displaystyle \lim_{n \to \infty} \bigg| \dfrac{2^{2n+3}}{(2n+3)!} \cdot \dfrac{(2n+1)!}{2^{2n+1}} \bigg|$

$\displaystyle \lim_{n \to \infty} \dfrac{4}{(2n+3)(2n+2)} = 0$

 October 30th, 2017, 08:53 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,658 Thanks: 2635 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}{k-1} \frac{2^k}{(k-2)!} \frac1{k^2} \\ &&&= \frac{k}{2(k-1)} \frac{2^{k+1}}{(k-2)!} \frac1{k^2} &= 2^4 \frac{k}{2(k-1)} \frac{2^{k-3}}{(k-2)!} \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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