Show that an Infinite Series is (absolutely) Convergent

Jan 2016
25
0
United Kingdom
Hello,

New to this topic after missing it at uni due to sickness! Would anyone be able to help me to do this question as im not really sure how to prove that something is convergent for all values of X.





Photo 10-02-2016, 08 21 17.jpg
[I do understand how to show that a series converges for a certain value of X]
 
Dec 2013
2,002
757
Colombia
For any real \(\displaystyle x\), pick \(\displaystyle N \in \mathbb N\) such that \(\displaystyle N > x\) and break the sum into two parts:

\(\displaystyle \sum \limits_{n=0}^\infty {x^n \over n!} = \sum \limits_{n=0}^{N-1} {x^n \over n!} + \sum \limits_{n=N}^\infty {x^n \over n!} < \sum \limits_{n=0}^{N-1} {x^n \over n!} + {x^N \over N!} \sum \limits_{n=0}^\infty {x^n \over "^n} \)​

The first term is a finite sum and the summation in the second term is a convergent geometric series.
 
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Dec 2013
2,002
757
Colombia
Oops, typo:
For any real \(\displaystyle x\), pick \(\displaystyle N \in \mathbb N\) such that \(\displaystyle N > x\) and break the sum into two parts:

\(\displaystyle \sum \limits_{n=0}^\infty {x^n \over n!} = \sum \limits_{n=0}^{N-1} {x^n \over n!} + \sum \limits_{n=N}^\infty {x^n \over n!} < \sum \limits_{n=0}^{N-1} {x^n \over n!} + {x^N \over N!} \sum \limits_{n=0}^\infty {x^n \over N^n} \)​

The first term is a finite sum and the summation in the second term is a convergent geometric series.
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
ratio test to determine the interval of convergence ...

$\displaystyle \lim_{n \to \infty} \bigg|\dfrac{a_{n+1}}{a_n}\bigg| < 1$

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

$\displaystyle |x| \cdot \lim_{n \to \infty} \dfrac{1}{n+1} < 1$

$|x| \cdot 0 < 1$ for all $x \in \mathbb{R}$ ... interval of convergence $(-\infty,\infty)$