#### To determine

Whether the series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is
∑n=1∞ln(n2+12n2+1)
.

Here,
an=ln(n2+12n2+1)
.

**Theorem used:** Series test for Divergence

If
limn→∞an
does not exist or
limn→∞an≠0
, then the series
∑n=1∞an
is divergent.

**Calculation:**

Obtain the limit of the sequence (the value of the term
an
as *n* tends to infinity).

limn→∞an=limn→∞(ln(n2+12n2+1))=ln(limn→∞(n2+12n2+1))
.

Divide the numerator and the denominator by its highest power.

limn→∞(ln(n2+12n2+1))=ln(limn→∞(n2+1n22n2+1n2))=ln(limn→∞(n2n2+1n22n2n2+1n2))=ln(limn→∞(1+1n22+1n2))=ln(1+limn→∞1n22+limn→∞1n2)

Apply infinity property
limn→∞(kna)=0
and simplify the terms of the expressions.

limn→∞(ln(n2+12n2+1))=ln(1+02+0)=ln(12)≈−0.6931

Thus, the limit of the sequence is
limn→∞(ln(n2+12n2+1))=ln(12)
.

Since
limn→∞(ln(n2+12n2+1))≠0
and by using the Theorem (Series test for Divergence), the series
∑n=1∞ln(n2+12n2+1)
is divergent.

Therefore, the series is divergent.