(a)

#### To determine

**To find:** The partial sum of the series and guess a formula for {sn}

#### Answer

The partial sums s1,s2,s3,s4 are 12,56,2324 and 119120 and the formula sn is (n+1)!−1(n+1)!.

#### Explanation

**Given:**

The series is ∑n=1∞n(n+1)!.

**Calculation:**

Consider the nth partial sum of the given series is ∑n=nn(n+1)!.

Obtain the sum of s1

s1=1(1+1)!=12!=12

Thus, the first partial sum is, s1=12.

Obtain the partial sum of s2.

s2=1(1+1)!+2(2+1)!=12!+23!=12+13=3+26

That is, s2=56.

Thus, the second partial sum is, s2=(2+1)!−1(2+1)!.

Obtain the partial sum of s3.

s3=1(1+1)!+2(2+1)!+3(3+1)!=12+13+324=12+8+324s3=2324

That is,

s3=24−124=(3+1)!−1(3+1)!

Thus, the first partial sum is, s3=(3+1)!−1(3+1)!.

Obtain the partial sum of s4.

s4=1(1+1)!+2(2+1)!+3(3+1)!+4(4+1)!=(1(1+1)!+2(2+1)!+3(3+1)!)+45!=2324+130=115+4120

That is, s4=119120.

Thus, the first partial sum is, s4=(4+1)!−1(4+1)!.

Therefore, the partial sums s1,s2,s3,s4 are 12,56,2324 and 119120.

Hence, it can be concluded that the formula sn is (n+1)!−1(n+1)!

(b)

#### To determine

**To prove:** The formula for sn is (n+1)!−1(n+1)!.

#### Explanation

**Proof:**

To prove that sn=(n+1)!−1(n+1)! for all *n* by induction.

*Base case:* n=1

To prove that the claim is true when n=1.

s1=12=2!−12!

Therefore, the claim is true when n=1.

*Induction hypothesis:* n=k

Assume that the claim is true when n=k.

sk=(k+1)!−1(k+1)!

*Induction step:* n=k+1

To prove that the claim is true when n=k+1.

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

sk+1=∑k=1n+1(k+1)!−1(k+1)!=sk+k+1(k+2)!=(k+1)!−1(k+1)!+k+1(k+2)(k+1)! (∵sk=(k+1)!−1(k+1)!)=(k+2)(k+1)!−(k+2)+k+1(k+2)(k+1)!

That is, sk+1=(k+2)!−1(k+2)!

Hence, by induction, the result sn=(n+1)!−1(n+1)! is true for all *n*.

(c)

#### To determine

**To show:** The infinite series ∑n=∞n(n+1)! is convergent and to find the sum of the series.

#### Explanation

**Proof:**

From part (a), sn=(n+1)!−1(n+1)!.

∑n=1∞n(n+1)!=limn→∞sn=limn→∞(n+1)!−1(n+1)!=limn→∞(n+1)!(n+1)!−limn→∞1(n+1)!=1−1∞=1

Since limit exist, it can be concluded that the given series is convergent.

Hence, the series is convergent to the limit 1.