To find the interval of convergence for the series [tex]\(\sum_{n=1}^{\infty} \frac{x^{11n+1}}{n!}\)[/tex], we will employ the ratio test. The ratio test involves examining the limit of the ratio of successive terms in the series.
### Step-by-Step Solution:
1. Identify the terms of the series:
The general term of the series is:
[tex]\[
a_n = \frac{x^{11n + 1}}{n!}
\][/tex]
2. Apply the Ratio Test:
Consider the ratio of the [tex]\((n+1)\)[/tex]-th term to the [tex]\(n\)[/tex]-th term:
[tex]\[
\text{Ratio} = \left| \frac{a_{n+1}}{a_n} \right|
\][/tex]
First, compute [tex]\(a_{n+1}\)[/tex]:
[tex]\[
a_{n+1} = \frac{x^{11(n+1) + 1}}{(n+1)!} = \frac{x^{11n + 12}}{(n+1)!}
\][/tex]
Then, find the ratio:
[tex]\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{x^{11n + 12}}{(n+1)!}}{\frac{x^{11n + 1}}{n!}} \right| = \left| \frac{x^{11n + 12} \cdot n!}{x^{11n + 1} \cdot (n + 1)!} \right|
\][/tex]
Simplify the ratio:
[tex]\[
= \left| \frac{x^{11n + 12}}{x^{11n + 1}} \cdot \frac{1}{n + 1} \right|
= \left| x^{11} \cdot \frac{1}{n + 1} \right|
\][/tex]
3. Take the limit as [tex]\(n\)[/tex] approaches infinity:
[tex]\[
\lim_{n \to \infty} \left| x^{11} \cdot \frac{1}{n + 1} \right| = \left| x^{11} \right| \cdot \lim_{n \to \infty} \frac{1}{n + 1} = \left| x^{11} \right| \cdot 0 = 0
\][/tex]
According to the ratio test, if this limit is less than 1, the series converges. Since:
[tex]\[
0 < 1
\][/tex]
The series converges for all values of [tex]\(x\)[/tex].
### Conclusion:
The above computations show that the ratio test yields a limit of 0, which is always less than 1 for any [tex]\(x\)[/tex]. Therefore, the series converges for all [tex]\(x\)[/tex].
So, the interval of convergence is:
[tex]\[
-\infty < x < \infty
\][/tex]
Thus, the radius of convergence is infinite, and the interval of convergence can be represented as:
[tex]\[
\boxed{-\infty < x < \infty}
\][/tex]
