Solve 1/0!+1/1!+1/2!+1/3!+frac{1}{4!}=

$$\frac{ 1 }{ 0 ! } + \frac{ 1 }{ 1 ! } + \frac{ 1 }{ 2 ! } + \frac{ 1 }{ 3 ! } + \frac{ 1 }{ 4 ! } =$$

Evaluate

$$\frac{65}{24}\approx 2.708333333$$

Solution Steps

The factorial of 0 is 1.

$$\frac{1}{1}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}$$

Anything divided by one gives itself.

$$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}$$

The factorial of 1 is 1.

$$1+\frac{1}{1}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}$$

Anything divided by one gives itself.

$$1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}$$

Add 1 and 1 to get 2.

$$2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}$$

The factorial of 2 is 2.

$$2+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}$$

Convert 2 to fraction \frac{4}{2}.

$$\frac{4}{2}+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}$$

Since \frac{4}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{4+1}{2}+\frac{1}{3!}+\frac{1}{4!}$$

Add 4 and 1 to get 5.

$$\frac{5}{2}+\frac{1}{3!}+\frac{1}{4!}$$

The factorial of 3 is 6.

$$\frac{5}{2}+\frac{1}{6}+\frac{1}{4!}$$

Least common multiple of 2 and 6 is 6. Convert \frac{5}{2} and \frac{1}{6} to fractions with denominator 6.

$$\frac{15}{6}+\frac{1}{6}+\frac{1}{4!}$$

Since \frac{15}{6} and \frac{1}{6} have the same denominator, add them by adding their numerators.

$$\frac{15+1}{6}+\frac{1}{4!}$$

Add 15 and 1 to get 16.

$$\frac{16}{6}+\frac{1}{4!}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$\frac{8}{3}+\frac{1}{4!}$$

The factorial of 4 is 24.

$$\frac{8}{3}+\frac{1}{24}$$

Least common multiple of 3 and 24 is 24. Convert \frac{8}{3} and \frac{1}{24} to fractions with denominator 24.

$$\frac{64}{24}+\frac{1}{24}$$

Since \frac{64}{24} and \frac{1}{24} have the same denominator, add them by adding their numerators.

$$\frac{64+1}{24}$$

Add 64 and 1 to get 65.

$$\frac{65}{24}$$

Factor

$$\frac{5 \cdot 13}{2 ^ {3} \cdot 3} = 2\frac{17}{24} = 2.7083333333333335$$