#### To determine

Whether the series is convergent or divergent.

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is
∑n=1∞an=∑n=1∞25n−1
.

**Result used**:

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

(2) The function
f(x)
is decreasing function if
f′(x)<0

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

**Definition used:**

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

**Calculation:**

Consider the function from given series
25x−1
.

The derivative of the function is obtained as follows,

Apply the chain rule (3) stated above,

f′(x)=2[(−1)(5x−1)−1−1⋅ddx(5x−1)]=2[−(5x−1)−2⋅(5)]=−10(5x−1)2<0

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 (1), the given series is divergent if the improper integral
∫1∞25x−1dx
is divergent.

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

Compute the limit of
∫1∞25x−1dx
.

∫1∞25x−1dx= limt→∞[∫1t25x−1dx]= limt→∞ [25ln(5x−1)]1t=limt→∞ (25[ln(5x−1)]1t)

Apply the lower limit and upper limit,

∫1∞25x−1dx=limt→∞(25[ln(5t−1)−ln(5−1)])=25[ln(∞)−ln(4)]=∞

Since the limit does not exist, it can be concluded that the improper integral is divergent.

Thus, the series is divergent.