#### To determine

Whether the series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is convergent to the sum ee−1.

#### Explanation

**Given:**

The series is ∑n=1∞(1en+1n(n+1)).

**Result used:**

(1) The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is convergent if |r|<1 and its sum is a1−r, where *a* is the first term and *r* is the common ratio of the series.

(2) If limn→∞sn=L, then ∑n=1∞an=L.

**Calculation:**

The given series can be written as follows,

∑n=1∞(1en+1n(n+1))=∑n=1∞1en+∑n=1∞1n(n+1)

=∑n=1∞(1e)n+∑n=1∞1n(n+1) (1)

Consider the series, ∑n=1∞(1e)n=(1e)+(1e)2+(1e)3+⋯.

Clearly, it is geometric series with the first term of the series, a=1e and the common ratio of the series is,

r=(1e)21e=1e

The absolute value of *r* is,

|r|=|1e|=1e<1

Since |r|<1 and by the Result (1) stated above, the series is convergent.

Thus, the series ∑n=1∞(1e)n is convergent.

Obtain the sum of the series.

∑n=1∞(1e)n=1e1−1e=1ee−1e=1e⋅ee−1=1e−1

Therefore, the sum of the series ∑n=1∞(1e)n=1e−1. (2)

Consider the series, ∑n=1∞1n(n+1).

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

1n(n+1)=a0n+a1n+1 (3)

Multiply the equation by n(n+1),

n(n+1)n(n+1)=a0n(n+1)n+a1n(n+1)n+11=a0(n+1)+a1n1=(a0+a1)n+a0

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

a0+a1=0

a0=1

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

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

1n(n+1)=1n−1n+1

Therefore, the series is ∑n=1∞1n(n+1)=∑n=1∞(1n−1n+1).

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

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

Take limit on both sides,

limn→∞sn=limn→∞(1−1n+1)=1−1∞=1−0=1

Since limn→∞sn=1, by the Result (2) stated above, ∑n=1∞1n(n+1)=1.

Therefore, the series convergent and ∑n=1∞1n(n+1)=1. (4)

Note that, the sum of two convergent series is convergent.

Since both the series ∑n=1∞(1e)n and ∑n=1∞1n(n+1) are convergent, then their sum ∑n=1∞(1en+1n(n+1)) is convergent.

Obtain the sum of the series ∑n=1∞(1en+1n(n+1)).

Substitute equation (2) and (4) in equation (1),

∑n=1∞(1en+1n(n+1))=1e−1+1=1e−1+e−1e−1=1+e−1e−1=ee−1

Thus, the sum of the series is ∑n=1∞(1en+1n(n+1))=ee−1.

The series is convergent to the sum ee−1.