Question

What is the interval of convergence of $\sum\limits_{}^{} {\dfrac{{{x^n}}}{{n!}}}$?

Answer 1

To find the interval of convergence we will first check whether the given function is convergent or divergent. For that we will use the ratio test. Then we will evaluate the limit. If the limit we get is less than one and independent of $x$, then the series will be convergent.

Complete step by step answer:
In a positive term series $\sum\limits_{}^{} {{u_n}}$, if $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{u_{n + 1}}}}{{{u_n}}} = \lambda$, then the series converges for $\lambda < 1$ and diverges for $\lambda > 1$.
Now we will take the absolute values and apply the ratio test.
We have;
$|{u_{n + 1}}| = \dfrac{{|x{|^{n + 1}}}}{{\left( {n + 1} \right)!}}$
$|{u_n}| = \dfrac{{|x{|^n}}}{{n!}}$
So, we have;
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{|{u_{n + 1}}|}}{{|{u_n}|}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\dfrac{{|x{|^{n + 1}}}}{{\left( {n + 1} \right)!}}} \right)}}{{\left( {\dfrac{{|x{|^n}}}{{n!}}} \right)}}$
Further solving we get;
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{|{u_{n + 1}}|}}{{|{u_n}|}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\dfrac{{|x| \times |x{|^n}}}{{\left( {n + 1} \right)n!}}} \right)}}{{\left( {\dfrac{{|x{|^n}}}{{n!}}} \right)}}$
Cancelling the terms in the numerator and the denominator, we get;
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{|{u_{n + 1}}|}}{{|{u_n}|}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{|x|}}{{n + 1}}$
Putting the value, we get;
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{|{u_{n + 1}}|}}{{|{u_n}|}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{|x|}}{{\infty + 1}}$
Using the concept that any number divided by infinity is equal to zero. We get;
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{|{u_{n + 1}}|}}{{|{u_n}|}} = 0$
Now, the value of this limit i.e., zero is less than one and also independent of $x$. So, the given series converges for all values of $x$.
Hence the interval of convergence of is $- \infty < x < \infty$.

Note:
The given series is convergent and it has a value equal to ${e^x}$. One point to note is that a sequence which is monotonic and bounded is convergent. A sequence ${a_n}$ is said to be bounded if there exists a number $k$ such that ${a_n} < k$ for every number $n$.