#### To determine

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

#### Answer

The series is convergent to the sum 12.

#### Explanation

**Given:**

The series is ∑n=4∞1n−1n+1.

Here, an=1n−1n+1.

**Result used:**

If the limit of the partial sums exists and limn→∞sn=L, then the series convergent and its sum is ∑n=1∞an=L.

**Calculation:**

Obtain the limit of the partial sums.

Let sn be the *n*th partial sum of the series ∑n=4∞1n−1n+1. Then,

sn=∑k=4n1k−1k+1=(14−14+1)+(15−15+1)+(16−16+1)+⋯+(1n−1n+1)=(12−15)+(15−16)+(16−17)+⋯+(1n−1n+1)

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

Therefore, sn=12−1n+1.

Take limits on both sides,

limn→∞sn=limn→∞(12−1n+1)=limn→∞12−limn→∞1n+1=12−0=12

Since limn→∞sn=12 and by the Result stated above, the series ∑n=4∞1n−1n+1 is convergent and its sum is ∑n=4∞1n−1n+1=12.

Therefore, the series convergent to the sum 12.