#### To determine

**To Find:** The sequence an and sum of the series ∑n=1∞an.

#### Answer

The sequence is an={0if n=12n2+nif n>1.

The sum of the series is ∑n=1∞an=1.

#### Explanation

The sequence is an={0if n=12n2+nif n>1.

The sum of the series is ∑n=1∞an=1.

**Given:**

The *n*th partial sum of the series is sn=n−1n+1. (1)

**Calculation:**

Obtain the sequence an.

For n=1, a1=s1.

Substitute 1 for *n* in equation (1),

s1=1−11+1=01=0

Thus, the first term of the sequence is a1=0.

For n>1,

sn=a1+a2+a3+⋯+an−1︸sn−1+an=sn−1+an

Therefore, an=sn−sn−1. (2)

Substitute n−1 for *n* in equation (1),

sn−1=(n−1)−1(n−1)+1=n−1−1n−1+1=n−2n

Therefore, the (n−1)th partial sum of the series is sn−1=n−2n. (3)

Substitute equations (1) and (3) in equation (2),

an=n(n−1)n(n+1)−(n−2)(n+1)n(n+1)=n(n−1)−(n−2)(n+1)n(n+1)=(n2−n)−(n2−2n+n−2)n2+n=2n2+n

Therefore, the sequence is an={0if n=12n2+nif n>1.

Obtain the sum of the series ∑n=1∞an.

∑n=1∞an=limn→∞(∑k=1nak)=limn→∞sn=limn→∞n−1n+1

Divide the numerator and the denominator by *n*.

∑n=1∞an=limn→∞n−1nn+1n=limn→∞1−1n1+1n=limn→∞1−limn→∞1nlimn→∞1+limn→∞1n=1−limn→∞1n1+limn→∞1n

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

∑n=1∞an=1−01+0=11=1

Therefore, the sum of the series is ∑n=1∞an=1.