Problem 6
Question
Find the first five terms of the sequence of partial sums. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !} $$
Step-by-Step Solution
Verified Answer
The first five terms of the sequence of partial sums are 1, 1/2, 2/3, 9/24, and 33/120.
1Step 1: 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 \((-1)^{1+1}/1! = 1/1 = 1\)
2Step 2: Identifying the Second Term
Plug n=2 into the formula to get the second term: \((-1)^{2+1}/2! = -1/2\)
3Step 3: Identifying the Third Term
Plug n=3 into the formula to get the third term: \((-1)^{3+1}/3! = 1/6\)
4Step 4: Identifying the Fourth Term
Plug n=4 into the formula to get the fourth term: \((-1)^{4+1}/4! = -1/24\)
5Step 5: Identifying the Fifth Term
Plug n=5 into the formula to get the fifth term: \((-1)^{5+1}/5! = 1/120\)
6Step 6: 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
Other exercises in this chapter
Problem 5
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}} $$
View solution Problem 5
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{3}{2 x-1}, \quad c=0 $$
View solution Problem 6
Write the first five terms of the sequence. \(a_{n}=(-1)^{n+1}\left(\frac{2}{n}\right)\)
View solution Problem 6
Use the Integral Test to determine the convergence or divergence of the series. $$ \frac{\ln 2}{2}+\frac{\ln 3}{3}+\frac{\ln 4}{4}+\frac{\ln 5}{5}+\frac{\ln 6}{
View solution