#### 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 32.

#### Explanation

**Given:**

The series is ∑n=2∞2n2−1.

Here, an=2n2−1.

**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:**

Consider, ∑n=2∞2n2−1.

∑n=2∞2n2−1=∑n=2∞2(n−1)(n+1) (1)

Obtain the partial fraction of 2(n−1)(n+1).

2(n−1)(n+1)=a0n−1+a1n+1 (2)

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

2(n−1)(n+1)(n−1)(n+1)=a0(n−1)(n+1)n−1+a1(n−1)(n+1)n+12=a0(n+1)+a1(n−1)2=a0n+a0+a1n−a12=(a0+a1)n+(a0−a1)

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

a0+a1=0a0−a1=2

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

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

2(n−1)(n+1)=1n−1+−1n+1=1n−1−1n+1

Therefore, the partial fraction is 2(n−1)(n+1)=1n−1−1n+1. (3)

Substitute equation (3) in equation (1), the series is ∑n=2∞1n(n+1)=∑n=2∞(1n−1−1n+1).

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

sn=∑k=2n(1k−1−1k+1)=(12−1−12+1)+(13−1−13+1)+⋯+(1(n−1)−1−1(n−1)+1)+(1n−1−1n+1)=(11−13)+(12−14)+(13−15)+⋯+(1n−1−1−1n−1+1)+(1n−1−1n+1)=(1−13)+(12−14)+(13−15)+⋯+(1n−2−1n)+(1n−1−1n+1)

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

Therefore, sn=1+12−1n−1n−1

Take limit on both sides,

limn→∞sn=limn→∞(1+12−1n−1n−1)=limn→∞(1)+limn→∞(12)−limn→∞(1n)−limn→∞(1n−1)=1+12−0+0 [∵limn→∞(1n)=0, limn→∞(1n−1)=0]=32

Since limn→∞sn=32, by the Result stated above, the series ∑n=2∞2n2−1 is convergent and its sum is ∑n=2∞2n2−1=32.

Therefore, the series convergent to the sum 32.