#### 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.

**Derivative rule:** Quotient Rule

If
f1(x)
and
f2(x)
are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2(x)]2

**Given:**

The series is
∑n=1∞an=∑n=1∞nn2+1
.

**Definition used:**

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

**Calculation:**

Consider the function from given series
xx2+1
.

The derivative of the function is obtained as follows,

f′(x)=(x2+1)ddx(x)−xddx(x2+1)(x2+1)2=(x2+1)(1)−x(2x)(x2+1)2=x2+1−2x2(x2+1)2=−(x2−1(x2+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 (1), the series is divergent if the improper integral
∫1∞xx2+1dx
is divergent.

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

Compute
∫1∞xx2+1dx
.

∫1∞xx2+1dx= limt→∞[∫1txx2+1dx]= limt→∞ [12ln(x2+1)]1t=limt→∞12[ln(t2+1)−ln(1+1)]

Apply the limit
t→∞
.

∫1∞xx2+1dx= 12(∞)=∞

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

Thus, the series is divergent.