#### To determine

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

#### Answer

The interval of convergence is (−3,−1) and the sum of the series is −x+2x+1.

#### Explanation

**Given:**

The series is ∑n=1∞(x+2)n.

**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=1∞(x+2)n=(x+2)+(x+2)2+(x+2)3+(x+2)4+⋯

Clearly, it is geometric series with first term of the series is a=x+2 and common ratio is r=x+2.

Use the Result stated above, the geometric series ∑n=1∞(x+2)n is converges if |r|<1.

|x+2|<1−1<x+2<1−1−2<x+2−2<1−2−3<x<−1

Thus, the series is converges when x∈(−3,−1).

Therefore, the interval of convergence is (−3,−1).

Obtain the sum of the geometric series.

Since a=x+2 and r=x+2, the sum of the series is computed as follows.

∑n=1∞(x+2)n=x+21−(x+2)=x+21−x−2=x+2−x−1=−x+2x+1

Therefore, the sum of the series is −x+2x+1.