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I am trying to find the limit of the sequence that is represented as following.

$$\sum_{i=0}^\infty \frac{{(-1)}^n n! {x}^n}{10^n}$$

Given its an alternating series I tried to prove that $|\frac{n! {x}^n}{10^n}|$ is convergent for which I tried ratio test which is coming out as

$\frac{T_{n+1}}{T_{n}} = \frac{x(n+1)}{10}$

and Raab test $n (\frac{T_{n}}{T_{n+1}} - 1)$ as

$n(\frac{10}{x(n+1)}-1)$

Both the ratio and Raab's test are kind of coming out as approaching $\infty$. I think I am missing something here, a little help will be deeply appreciated.

Thanks Much

Sid

Sid111Math
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1 Answers1

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For a series $\sum a_n$ to converge it is necessary that $a_n \to 0$. Let us show that $\frac {n! |x|^{n}} {10^{n}} \to \infty$ for every $x \neq 0$. That would show that the series converges only when $x=0$.

Choose a positive integer $N$ such that $\frac {|x|N} {10} >2$. Since $n! >(N+1)(N+2)...(n) >N^{n-N}$ we see that $\frac {n! |x|^{n}} {10^{n}} \to \infty >\frac {N^{n-N} |x|^{n}} {10^{n}} \to \infty$ because $\frac {|x|N} {10} >2$ and $2^{n} \to \infty$.

Kavi Rama Murthy
  • 311,013
  • thanks, it kind of shows how it diverge for all $x \neq 0$ but is there a way we can prove it by ratio, raab, logarithmic or cauchy integral test. – Sid111Math Dec 30 '19 at 03:22
  • @Sid111Math When you test $\sum a_n$ for convergence it is recommended that you first check if $a_n to 0$. The test are to be used only after making sure that $a_n \to 0$. – Kavi Rama Murthy Dec 30 '19 at 05:14