#### 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 14.

#### Explanation

**Given:**

The series is ∑n=2∞1n3−n.

Here, an=1n3−n.

**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:**

Consider the series, ∑n=2∞1n3−n. (1)

Obtain the partial fraction of 1n3−n.

1n3−n=1n(n2−1)=1n(n+1)(n−1)

=a1n(n−1)+a2n(n+1) (2)

Multiply the equation by (n+1)(n−1) and simplify the terms,

(n+1)(n−1)n3−n=a1(n+1)(n−1)n(n−1)+a2(n+1)(n−1)n(n+1)1n=a1(n+1)n+a2(n−1)n1=a1(n+1)+a2(n−1)1=(a1+a2)n+(a1−a2)

Equate the coefficient of *n* and the constant term on both sides,

a1+a2=0a1−a2=1

Solve the above equations and obtain a1=12and a2=−12.

Substitute the values a1=12and a2=−12 in equation (2),

1n3−n=12n(n−1)+−12n(n+1)=12n(n−1)−12n(n+1)=12(1n(n−1)−1n(n+1))

Therefore, the partial fraction is 1n3−n=12(1n(n−1)−1n(n+1)). (3)

Substitute equation (3) in equation (1), then the series is

∑n=2∞1n3−n=∑n=2∞12(1n(n−1)−1n(n+1))=12⋅∑n=2∞(1n(n−1)−1n(n+1))=12⋅[∑n=2∞(1n(n−1)−1n(n+1))]

Therefore, the series is ∑n=2∞1n3−n=12⋅[∑n=2∞(1n(n−1)−1n(n+1))].

Let sn be the *n*th partial sum of the series ∑n=2∞1n3−n. Then,

sn=∑k=2n1k3−k=12⋅[∑k=2n1k(k−1)−1k(k+1)]=12⋅[(12(2−1)−12(2+1))+(13(3−1)−13(3+1))+(14(4−1)−14(4+1))+⋯+ (1(n−1)(n−1−1)−1(n−1)(n−1+1))+(1n(n−1)−1n(n+1))]=12⋅[(12⋅1−12⋅3)+(13⋅2−13⋅4)+(14⋅3−14⋅5)+⋯+(1(n−1)(n−2)−1n(n−1)) +(1n(n−1)−1n(n+1))]

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

Therefore, the series becomes,

sn=12[12−1n(n+1)]=14−12n(n+1)

Take limit on both sides,

limn→∞sn=limn→∞(14−12n(n+1))=14−1∞=14−0=14

Since limn→∞sn=14 and by the Result stated above, the series ∑n=2∞1n3−n is convergent and its sum is ∑n=2∞1n3−n=14.

Therefore, the series convergent to the sum 14.