Question

Find the first five terms of the sequence of partial sums. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !} $$

Short answer

The first five terms of the sequence of partial sums are 1, 1/2, 2/3, 9/24, and 33/120.

Step-by-step solution

Identifying the First Term

The given formula is \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}\). Therefore, when you substitute n=1 into the formula, the first term of the series becomes \((-1)^{1+1}/1! = 1/1 = 1\)

Identifying the Second Term

Plug n=2 into the formula to get the second term: \((-1)^{2+1}/2! = -1/2\)

Identifying the Third Term

Plug n=3 into the formula to get the third term: \((-1)^{3+1}/3! = 1/6\)

Identifying the Fourth Term

Plug n=4 into the formula to get the fourth term: \((-1)^{4+1}/4! = -1/24\)

Identifying the Fifth Term

Plug n=5 into the formula to get the fifth term: \((-1)^{5+1}/5! = 1/120\)

Creating the Sequence of Partial Sums

Now, add together the terms obtained to form the first five terms of the partial sums. The partial sums are 1, 1-1/2, 1-1/2+1/6, 1-1/2+1/6-1/24, 1-1/2+1/6-1/24+1/120, simplifying this sequence of partial sums you will get: 1, 1/2, 2/3, 9/24, 33/120