#### To determine

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

#### Answer

The interval of convergence is (−∞,0) and the sum of the series is 11−ex.

#### Explanation

**Given:**

The series is ∑n=0∞enx.

**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 written as follows,

∑n=0∞enx=∑n=0∞(ex)n=(ex)0+(ex)1+(ex)2+(ex)3+⋯=1+(ex)+(ex)2+(ex)3+⋯

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

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

|ex|<1ex<1

Take natural logarithm on both the sides and use the result ln(ex)=x and ln1=0,

ln(ex)<ln1x<0

Thus, the series converges if x∈(−∞,0).

Therefore, the interval of convergence is (−∞,0).

Obtain the sum of the geometric series.

Since a=1 and r=ex, the sum of the series is, ∑n=0∞enx=11−ex.

Therefore, the sum of the series is 11−ex.