#### To determine

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

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is ∑n=1∞lnnn+1.

Here, an=lnnn+1.

**Result used:**

If the limit of the partial sums limn→∞sn does not exist, then the series is divergent.

**Calculation:**

Consider the series, ∑n=1∞lnnn+1.

Let sn be the *n*th partial sum of the series ∑n=1∞lnnn+1. Then,

sn=∑k=1nlnkk+1=ln(11+1)+ln(22+1)+ln(33+1)+⋯+ln(n−1n−1+1)+ln(nn+1)=ln(12)+ln(23)+ln(34)+⋯+ln(n−1n)+ln(nn+1)

Apply the Logarithm quotient rule ln(xy)=ln(x)−ln(y),

sn=(ln1−ln2)+(ln2−ln3)+(ln3−ln4)+⋯+(ln(n−1)−ln(n))+(ln(n)−ln(n+1))

Here, it is noticed that the terms are cancelled (telescoping sum) except the first term of the first parentheses and the second term of the last parentheses.

sn=ln1−ln(n+1)=0−ln(n+1) [∵ln1=0]=−ln(n+1)

Therefore, sn=−ln(n+1).

Take limit on both sides,

limn→∞sn=limn→∞(−ln(n+1))=−(limn→∞(ln(n+1)))=−ln(∞)=−∞

Since limn→∞sn does not exist and by the Result stated above, the series ∑n=1∞lnnn+1 is divergent.

Therefore, the series divergent.