#### To determine

Whether the series is convergent or divergent.

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is
13+17+111+115+119+...
.

**Result used**:

If the function
f(x)
is continuous, positive and decreasing on
[1,∞)
and let
an=f(n)
, then the series
∑n=1∞an
is divergent if and only if the improper integral
∫1∞f(x)dx
is divergent.

**Chain rule:**
d[f(x)]ndx=n[f(x)]n−1f′(x)

**Definition used:**

The improper integral is divergent if the limit does not exist.

**Calculation:**

The given series can be expressed as follows,

13+17+111+115+119+...=1(4(1)−1)+1(4(2)−1)+1(4(3)−1)+1(4(4)−1)+1(4(5)−1)+...=∑n=1∞1(4n−1)=∑n=1∞(4n−1)−1

Consider the function from given series
(4x−1)−1
.

f′(x)=(−1)(4x−1)−1−1[d(4x−1)dx] (Chain rule)=−[(4x−1)−2(4)]=−[4(4x−1)2]

Since
f′(x)<0
, the given function is decreasing by using the Result (2).

Clearly, the function
f(x)
is continuous, positive and decreasing on
[1,∞)

Use the result stated above, the series is convergent if the improper integral
∫1∞1(4x−1)dx
is convergent.

Use the definition, the improper integral is convergent if the limit is exists.

Compute
∫1∞1(4x−1)dx
as shown below.

∫1∞1(4x−1)dx= limt→∞[∫1t1(4x−1)dx]

Obtain the integral
∫1t1(4x−1)dx
.

Consider the integral
∫1(4x−1)dx
.

Substitute
u=4x−1 and du=dx

∫1(4x−1)dx=∫14udu=14∫1udu=14lnu=14ln(4x−1)

Compute the boundaries,

∫1t1(4x−1)dx=[ln(4x−1)]1t=ln(4t−1)−ln(4(1)−1)=ln(4t−1)−ln(4−1)=ln(4t−1)−ln3

Therefore,
∫1t1(4x−1)dx=ln(4t−1)−ln3
.

From equation (1),

∫1∞1(4n−1)dx=limt→∞[ln(4t−1)−ln3]=[ln(∞)−ln(3)]=[∞−ln3]=∞

Since the limit does not exists, it can be conclude that the improper integral is divergent.

Therefore, the given series is divergent.