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October 29th, 2017, 10:00 AM   #1
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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?
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October 29th, 2017, 10:55 AM   #2
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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?
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October 30th, 2017, 05:44 AM   #3
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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)
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October 30th, 2017, 06:20 AM   #4
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Quote:
Originally Posted by Jaket1 View Post
... and is greater than 2^k/k! for all k>0
Not all $k > 0$. It's not true for $ k=1$ for example.
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October 30th, 2017, 09:25 AM   #5
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Quote:
Originally Posted by Jaket1 View Post
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 ...

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

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

$\displaystyle \lim_{n \to \infty} \dfrac{4}{(2n+3)(2n+2)} = 0$
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October 30th, 2017, 09:53 AM   #6
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\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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