#### To determine

Whether the series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is convergent to the sum
58
.

#### Explanation

**Given:**

The series is
13+29+127+281+1243+2719+⋯
.

**Definition used:**

If the sum of all terms of the series is finite, then the series is said to be convergent series.

**Result used:**

The geometric series
∑n=1∞arn−1 (or) a+ar+ar2+⋯
is convergent 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:**

The given series can be written as follows,

13+29+127+281+1243+2719+⋯=(13+127+1243+⋯)+(29+281+2719+⋯)
(1)

Here, both the series
13+127+1243+⋯ and 29+281+2719+⋯
are geometric series.

Obtain the sum of the geometric series
13+127+1243+⋯
if the series is convergent.

Consider
13+127+1243+⋯
.

Here, the first term of the series is,
a1=13
and the common ratio of the series is,

r1=12713=127⋅31=19

The absolute value of
r1
is,

|r1|=|19|=19=0.1111<1

Since
|r1|<1
and by the Result stated above, the geometric series
13+127+1243+⋯
is convergent.

The sum of the series is computed as follows,

a11−r1=131−19=1389=13⋅98=38

Therefore, the sum is
13+127+1243+⋯=38
(2)

Obtain the sum of the geometric series
29+281+2719+⋯
if the series is convergent.

Consider
29+281+2719+⋯
.

Here, the first term of the series is
a2=13
, and the common ratio of the series is,

r2=28129=281⋅92=19

The absolute value of
r2
is,

|r2|=|19|=19=0.1111<1

Since
|r2|<1
and by the Result stated above, the geometric series
29+281+2719+⋯
is convergent.

The sum of the series is as follows,

a21−r2=291−19=2989=29⋅98=14

Therefore, the sum is
29+281+2719+⋯=14
(3)

Substitute equation (2) and (3) in equation (1),

13+29+127+281+1243+2719+⋯=38+14=38+14⋅22=38+28=58

Since the sum of all terms of the given series
58
is finite and by using the definition, the given series is convergent

Therefore, the series is convergent to the sum
58
.