#### 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 116.

#### Explanation

**Given:**

The series is ∑n=1∞3n(n+3).

Here, an=3n(n+3).

**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=1∞3n(n+3). (1)

Obtain the partial fraction of 3n(n+3).

3n(n+3)=a0n+3+a1n (2)

Multiply the equation by n(n+3) and simplify the terms,

3n(n+3)=a0n+3+a1n3=a0n+a1(n+3)3=a0n+a1n+3a13=(a0+a1)n+(3a1)

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

a0+a1=03a1=3

Solve the above equations and obtain a0=1 and a1=−1.

Substitute 1 for a0 and -1 for a1 in equation (3),

3n(n+3)=1n+3+−1n=1n+3−1n

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

Substitute equation (3) in equation (1), the series is ∑n=1∞3n(n+3)=∑n=1∞(1n−1n+3).

Let sn be the *n*th partial sum of the series ∑n=1∞(1n−1n+3). Then,

sn=∑k=1n(1k−1k+3)={(11−11+3)+(12−12+3)+(13−13+3)+(14−14+3)+⋯+(1n−2−1n−2+3) +(1n−1−1n−2+3) +(1n−1n−1+3)}={(1−14)+(12−15)+(13−16)+(14−17)+⋯+(1n−2−1n+1) +(1n−1−1n+2) +(1n−1n+3)}

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

Therefore, sn=1+12+13−1n+1−1n+2−1n+3.

Take limit on both sides,

limn→∞sn=limn→∞(1+12+13−1n+1−1n+2−1n+3)=1+12+13−0−0−0=1⋅6+1⋅3+1⋅26=116

Since limn→∞sn=116, by the Result stated above, the series ∑n=1∞3n(n+3) is convergent and its sum is ∑n=1∞3n(n+3)=116.

Therefore, the series convergent to the sum 116.