#### To determine

Whether the series is convergent or divergent.

#### Answer

The series is convergent.

#### Explanation

**Given:**

The series is
∑n=1∞an=∑n=1∞n−3
.

**Definition used:**

The improper integral
∫abf(x)dx
is convergent if the corresponding limit exists.

**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
.

**Calculation:**

Consider the function from given series
x−3
.

The derivative of the function is obtained as follows,

f′(x)=(−3)x(−3−1)=−(3x−4)<0

Therefore, 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 series is convergent if the improper integral
∫1∞x−3dx
is convergent.

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

Compute
∫1∞x−3dx
as shown below.

∫1∞x−3dx=limt→∞[∫1tx−3dx]=limt→∞[x−2−2]1t=lim−t→∞12[t−2−1]

Apply the limit
t→∞
,

∫1∞x−3dx=−12[1∞−1]=−12[−1]=12

Since the limit exists, it can be concluded that the improper integral is convergent.

Thus, the series is convergent.