The Problem:

summation as 'n' goes from 2 to infinity of n^20/2^n .

The ans says it converges to 1/2 but cant see how it does this since the terms seem to be increasing in magnitude when you plug numbers in.

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- Sep 9th 2010, 02:30 AMheatlyInfinite Series
The Problem:

summation as 'n' goes from 2 to infinity of n^20/2^n .

The ans says it converges to 1/2 but cant see how it does this since the terms seem to be increasing in magnitude when you plug numbers in. - Sep 9th 2010, 02:38 AMyeKciM
$\displaystyle \displaystyle \sum _{n=2} ^{\infty} \frac {n^{20}}{2^n} $

so you have that your a_n is

$\displaystyle \displaystyle a_n = \frac {n^{20}}{2^n} $

now apply some of the tests for convergence ...

you can use Cauchy's test of convergence :

$\displaystyle \displaystyle q= \lim _{n\to \infty} \sqrt [n] {a_n} = \lim_{n \to \infty }\sqrt [n] { \frac {n^{20}}{2^n}} = \left\{\begin{matrix}

q<1 & converges \\

q>1 & diverges \\

q=1 & ???

\end{matrix}\right.$

or with D'Alambert's test for convergence :

$\displaystyle \displaystyle q= \lim_{n \to \infty } \frac {a_{n+1}}{a_n} =\lim_{n \to \infty } \frac {\frac {(n+1)^{20}}{2^{n+1}} }{\frac {n^{20}}{2^n}} = \left\{\begin{matrix}

q<1 & converges \\

q>1 & diverges \\

q=1 & ???

\end{matrix}\right.$

or with Raabe's test of convergence :

$\displaystyle \displaystyle q= \lim_{n \to \infty } n(\frac {a_{n}}{a_{n+1} }-1 ) =\lim_{n \to \infty } n ( \frac {\frac {n^{20}}{2^n} }{\frac {(n+1)^{20}}{2^{n+1}}}-1 ) = \left\{\begin{matrix}

q>1 & converges \\

q<1 & diverges \\

q=1 & ???

\end{matrix}\right.$

or you can show by the definition that sum converges or diverges :D (with partial sums .... ) - Sep 9th 2010, 03:30 AMmr fantastic
- Sep 9th 2010, 03:52 AMyeKciM
$\displaystyle \displaystyle \sum _{n=2} ^{\infty} \frac {n^{20}}{2^n} $

so based on D'Alambert test of convergence :

$\displaystyle \displaystyle q= \lim_{n \to \infty } \frac {a_{n+1}}{a_n} =\lim_{n \to \infty } \frac {\frac {(n+1)^{20}}{2^{n+1}} }{\frac {n^{20}}{2^n}} =\lim_{n \to \infty } \frac {\frac {(n+1)^{20}}{2^n\cdot 2}} {\frac {n^{20}}{2^n}} = \lim_{n \to \infty } \frac {n^{20 } + ... } { 2n^{20} } = \frac {1}{2} $

now because $\displaystyle q = \frac {1}{2} <1 $ based on D'Alambert test of convergence sum $\displaystyle \displaystyle \sum _{n=2} ^{\infty} \frac {n^{20}}{2^n} $ converges :D:D:D

maybe that's what ans was about :D