Chapter 9.4, Problem 59E

To determine

To find: To find whether the statement is true or false.

Answer to Problem 59E

The answer is $\frac{1}{2}$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum \frac{{(\sin x)}^n}{2^n{n}^2}}$ .

Calculation:

The given function is ${\displaystyle \sum \frac{{(\sin x)}^n}{2^n{n}^2}}$ .

  ${(\sin x)}^n\le 1$ for any value of $x$ . For any $n$ we have $\frac{{(\sin x)}^n}{2^n{n}^2}\le \frac{1}{2^n{n}^2}\le \frac{1}{n^2}$

  ${\displaystyle \sum \frac{{(\sin x)}^n}{2^n{n}^2}}<{\displaystyle \sum \frac{1}{2^n}}$

Since we know converges for all real numbers. The Series dominates ${\displaystyle \sum \frac{{(\sin x)}^n}{2^n{n}^2}}$ term by term, the latter series must also converge for all real numbers by direct comparison.

Since ${\displaystyle \sum \frac{2{(\sin x)}^n}{n!+3}}$ converges absolutely, Therefore, it converges.

The answer is E.