# Show that an Infinite Series is (absolutely) Convergent

#### thisuserhas

All major convergence tests are detailed here:
https://en.wikipedia.org/wiki/Convergence_tests
I generally start with the limit test and then move to more 'complicated' tests if it renders an inconclusive result.
Hint - for 1(b)
$$\displaystyle a_{n} = x^{n}/n!$$

#### Archie

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.

• 1 person

#### Archie

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
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 $(-\infty,\infty)$