#### To determine

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

#### Answer

The series is convergent and the sum is e−1.

#### Explanation

**Given:**

The series is ∑n=1∞(e1n−e1(n+1)).

Here, an=e1n−e1(n+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:**

Obtain the limit of the partial sums.

Let sn be the *n*th partial sum of the series ∑n=1∞(e1n−e1(n+1)). Then,

sn=∑k=1ne1k−e1(k+1)=(e11−e1(1+1))+(e12−e1(2+1))+(e13−e1(3+1))+⋯+(e1n−1−e1(n−1+1))+(e1n−e1(n+1))=(e1−e12)+(e12−e13)+(e13−e14)+⋯+(e1n−1−e1n)+(e1n−e1(n+1))

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

Therefore, sn=e1−e1(n+1).

Take limit on both sides,

limn→∞sn=limn→∞(e−e1(n+1))=e−e1∞=e−e0=e−1

Since limn→∞sn=e−1 and by the Result stated above, the series ∑n=1∞(e1n−e1(n+1)) is convergent and its sum is ∑n=1∞(e1n−e1(n+1))=e−1.

Therefore, the series is convergent and the sum is e−1.