#### To determine

Whether the series is convergent or divergent.

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is
∑n=1∞an=∑n=1∞n−0.3
.

**Result used:** Integral Test

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.

**Calculation:**

Consider the function from given series
x−0.3
.

The derivative of the function is obtained as follows,

f′(x)=(−0.3)x0.7=−(0.3x0.7)

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

Use the Result (1), the series is divergent if the improper integral
∫1∞x−0.3dx
is divergent.

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

Compute
∫1∞x−0.3dx
as shown below.

∫1∞x−0.3dx= limt→∞[∫1tx−0.3dx]= limt→∞ [x0.70.7]1t=limt→∞10.7[t0.7−1]

Apply the limit
t→∞
,

∫1∞x−0.3dx=10.7[∞−1]=∞

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

Therefore, the series is divergent.