#### To determine

**To find:** The values of *x* if the given series is convergent and its sum.

#### Answer

The interval of convergence is (−∞,∞) and the sum of the series is 33−sinx.

#### Explanation

**Given:**

The series is ∑n=0∞sinnx3n.

**Result used:**

The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is converges if |r|<1 and its sum is a1−r, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

Obtain the value of *x* (the interval of converges).

The given series can be expressed as follows,

∑n=0∞sinnx3n=∑n=0∞(sinx3)n=(sinx3)0+(sinx3)1+(sinx3)2+(sinx3)3+⋯=1+(sinx3)+(sinx3)2+(sinx3)3+⋯

Clearly, it is geometric series with first term of the series is a=1 and common ratio is r=sinx3.

Use the Result stated above, the geometric series ∑n=0∞sinnx3n is converges if |r|<1.

|sinx3|<1|sinx||3|<1 [By the properties of modulus, |ab|=|a||b|]|sinx|3<1

Multiply the above inequality by 3,

|sinx|3⋅3<1⋅3|sinx|<3

Note that, |sinx|<1 for all x then |sinx|<3 is true for all values of *x.*

That is, the series is converges if x∈(−∞,∞).

Therefore, the interval of convergence is (−∞,∞).

Obtain the sum of the geometric series.

Since a=1 and r=sinx3.

∑n=0∞sinnx3n=11−sinx3=133−sinx3=13−sinx3=33−sinx

Therefore, the sum of the series is 33−sinx.