#### To determine

**To show:** The given series is divergent.

#### Explanation

The series is ∑n=1∞ln(1+1n).

Here, an=ln(1+1n).

**Result used:**

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

**Proof:**

Rewrite the given series and expressed as follows,

∑n=1∞ln(1+1n)=∑n=1∞ln(1+1n)=∑n=1∞ln(nn+1n)=∑n=1∞ln(n+1n)

Let sn be the *n*th partial sum of the series ∑k=1n(ln(k+1)−ln(k)). Then,

sn=∑k=1nln(k+1k)=ln(1+11)+ln(2+12)+ln(3+13)+ ⋯+ln((n−1)+1n−1)+ln(n+1n)=ln(21)+ln(32)+ln(43)+ ⋯+ln(nn−1)+ln(n+1n)

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

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

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

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

Therefore, sn=ln(n+1).

Take limit on both the sides,

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

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

Hence the required proof is obtained.