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Claim: $a_n = \frac{n^5}{2^n}$ diverges

Let M be arbitrary

Then $$ \forall \; n \ge \text{max} \big\{ \big[ \sqrt[3]{M} \, \big] + 1 , 3 \big\} \\ n > \sqrt[3]{M} \\ \implies n^3 > M $$

And Since $$ \forall \; n \ge 3 \\ n^2 > 2^n $$

It is true that $$ \frac{n^5}{2^n} > \frac{n^5}{n^2} = n^3 > M $$

$ \therefore \; a_n $ diverges

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$n^2 > 2^n$ is false for large $n$ – Emanuele Paolini Mar 15 '13 at 17:52
ah right - put that in an answer and ill accept it – DanZimm Mar 15 '13 at 17:53
up vote 6 down vote accepted

The claim is false, and the part

And Since $$ \forall \; n \ge 3 \\ n^2 > 2^n $$

is erroneous.

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I think you should look at the graph of $y=x^5/2^x$ from $x=0$ to $30$ to get some visual understanding of this sequence:

enter image description here

so it converges to zero very quickly after a small bump of working out which of $x^5$ or $2^x$ dominates.

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$$a_n = \frac{n^5}{2^n}$$ $$a_{n+1} = \frac{(n+1)^5}{2^{n+1}}$$

Determine the converging radius


$$\frac{a_{n+1}}{a_n}=\frac{(\frac{n+1}{n})^5}{2}$$ $$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n})^5}{2}$$

for $n\ge7$ $$\frac{(1+\frac{1}{n})^5}{2} < 1$$

Or the radius of convergence < 1 for n >= 7

So the series converges

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Since both components of the fraction are going to infinity, you can apply L'Hôpital's rule.
Doing so a few times, you will get constant in numerator and $2^n$ in denominator, which is enough to conclude that the limit is equal to $0$.

The same is true for any integer power of $n$ in numerator, or in other words exponential function goes to infinity faster than power function.

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If you're familiar with computational mathematics, you'll know that exponential running time ($O(a^n)$, for $a>1$) has a much faster order of growth than polynomial time, $O(n^c)$, for $c>1$. And so your quotient must vanish to zero.

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