#### To determine

Whether the series is convergent or divergent.

#### Answer

The series is divergent.

#### Explanation

**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.

**Definition used:**

The improper integral
∫abf(x)dx
is divergent if the limit does not exist.

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

**Given:**

The series is
15+17+19+111+113+...
.

**Calculation:**

The given series can be written as follows,

15+17+19+111+113+...=1(2(1)+3)+1(2(2)+3)+1(2(3)+3)+1(2(4)+3)+1(2(5)+3)+...=∑n=1∞1(2n+3)=∑n=1∞(2n+3)−1

Consider the function from the above series
f(x)=(2x+3)−1
.

The derivative of the function is computed as follows,

f′(x)=(−1)(2x+3)−1−1[ddx(2x+3)] (by chain rule)=−[(2x+3)−2(2)]=−2(2x+3)−2=−[2(2x+3)2]

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

Use the result stated above, the given series is divergent if the improper integral
∫1∞1(2x+3)dx
is divergent.

By the definition, the improper integral is divergent if the limit does not exist.

Compute
∫1∞1(2x+3)dx
as shown below,

∫1∞1(2x+3)dx= limt→∞[∫1t1(2x+3)dx]=limt→∞[ln(2x+3)2]1t=limt→∞[ln(2t+3)−ln(2(1)+3)2]

Apply the limit
t→∞
,

∫1∞1(2x+3)dx=[ln(∞)−ln(5)2]=∞

Since the limit does not exist, it can be concluded that the given series is divergent.