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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. Tags converge, diverge, sequence, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sagicoh Real Analysis 6 December 20th, 2012 08:43 AM Yoyojam55 Calculus 3 June 15th, 2012 07:35 AM Rita James Real Analysis 4 January 23rd, 2012 11:03 PM hunnynut Calculus 2 March 24th, 2010 08:12 AM zeion Calculus 2 March 24th, 2010 08:00 AM

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