Question

find the indicated sum.
\\(\sum\limits_{i=1}^{5}\frac{(-1)^{i}}{(i + 1)!}\\)
\\(\sum\limits_{i=1}^{5}\frac{(-1)^{i}}{(i + 1)!}=\square\\) (type an integer or a simplified fraction.)

Explanation:

Step1: Expand the sum for i=1 to 5

Compute each term:

• i=1: $\frac{(-1)^1}{(1+1)!} = -\frac{1}{2!} = -\frac{1}{2}$
• i=2: $\frac{(-1)^2}{(2+1)!} = \frac{1}{3!} = \frac{1}{6}$
• i=3: $\frac{(-1)^3}{(3+1)!} = -\frac{1}{4!} = -\frac{1}{24}$
• i=4: $\frac{(-1)^4}{(4+1)!} = \frac{1}{5!} = \frac{1}{120}$
• i=5: $\frac{(-1)^5}{(5+1)!} = -\frac{1}{6!} = -\frac{1}{720}$

Step2: Sum the terms

Convert to common denominator 720:
$-\frac{360}{720} + \frac{120}{720} - \frac{30}{720} + \frac{6}{720} - \frac{1}{720} = \frac{-360+120-30+6-1}{720} = \frac{-265}{720}$

Step3: Simplify the fraction

Divide numerator and denominator by 5: $\frac{-53}{144}$

Answer:

$-\frac{53}{144}$