1. ## Convergence series ...

Hi all!

How would you do these questions?

a) Find the range of values of $x$ for convergence of series:

$x + \frac{2^4x^2}{2!}+\frac{3^4x^3}{3!}+\frac{4^4x^4}{ 4!} + ...$

b) Investigate convergence of series

$\frac{1}{1.2}+\frac{1}{2.2^2}+\frac{1}{3.2^3}+\fra c{1}{4.2^4}+ ...$

Thank you! :-)

2. Originally Posted by Enita
Hi all!

How would you do these questions?

a) Find the range of values of $x$ for convergence of series:

$x + \frac{2^4x^2}{2!}+\frac{3^4x^3}{3!}+\frac{4^4x^4}{ 4!} + ...$
The series is,
$\sum_{n=1}^{\infty} \frac{n^4 x^n}{n!}$
Find the limit of,
$\frac{a_{n+1}}{a_n}=\left|\frac{(n+1)^4}{(n+1)!} \cdot \frac{n!}{n^4}\right|=\left|\frac{n!}{n!(n+1)} \cdot \frac{(n+1)^4}{n^4}\right|=\frac{(n+1)^3}{n^4}$
Open,
$\frac{n^3+3n^2+3n+1}{n^4}$
The limit converges to zero.
Thus, this series only converges for all $x$.

3. Originally Posted by Enita

b) Investigate convergence of series

$\frac{1}{1.2}+\frac{1}{2.2^2}+\frac{1}{3.2^3}+\fra c{1}{4.2^4}+ ...$
You can see that,
$0\leq \frac{1}{1\cdot 2}+\frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+... \leq \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...$
This sequence is bounded by two convergent sequences.
Because,
$\frac{1}{2}+\frac{1}{2^2}+...$
Is a geometric series and it converges.

This is mine 43th Post!!!

4. thanks theperfecthacker!