Chapter 10.4, Problem 3E

To determine

The series that converges for all x.

Answer to Problem 3E

The series dominates ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}$ term by term, the latter series must also converge for all real numbers by the Direct Comparison Test.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}$

Formula used:

  $\frac{x^{3n}}{2n!+1}\le \frac{x^{3n}}{n!}=\frac{{\left(x^3\right)}^n}{n!}{\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}\le {\displaystyle \sum _{n=0}^{\infty }\frac{{\left(x^3\right)}^n}{n!}}$

Calculation:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}$ It has no negative terms.

For any n, $\frac{x^{3n}}{2n!+1}\le \frac{x^{3n}}{n!}=\frac{{\left(x^3\right)}^n}{n!}{\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}\le {\displaystyle \sum _{n=0}^{\infty }\frac{{\left(x^3\right)}^n}{n!}}$

Since,

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(x^3\right)}^n}{n!}}$ is a s series for $e^{x^3}$ , which converges to $e^{x^3}$ for all real numbers.

Since the series dominates ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}$ term by term, the latter series must also converge for all real numbers by the Direct Comparison Test.

Conclusion:

The series dominates ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{2n!+1}}$ term by term, the latter series must also converge for all real numbers by the Direct Comparison Test.