Solve e=1+1/1!+11/2!+1/3!=

$$e = 1 + \frac { 1 } { 1 ! } + 1 \frac { 1 } { 2 ! } + \frac { 1 } { 3 ! } =$$

Verify

Solution Steps

The factorial of 1 is 1.

$$e=1+\frac{1}{1}+1\times \frac{1}{2!}+\frac{1}{3!}$$

Anything divided by one gives itself.

$$e=1+1+1\times \frac{1}{2!}+\frac{1}{3!}$$

Add 1 and 1 to get 2.

$$e=2+1\times \frac{1}{2!}+\frac{1}{3!}$$

The factorial of 2 is 2.

$$e=2+1\times \frac{1}{2}+\frac{1}{3!}$$

Multiply 1 and \frac{1}{2} to get \frac{1}{2}.

$$e=2+\frac{1}{2}+\frac{1}{3!}$$

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

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

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

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

Add 4 and 1 to get 5.

$$e=\frac{5}{2}+\frac{1}{3!}$$

The factorial of 3 is 6.

$$e=\frac{5}{2}+\frac{1}{6}$$

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

$$e=\frac{15}{6}+\frac{1}{6}$$

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

$$e=\frac{15+1}{6}$$

Add 15 and 1 to get 16.

$$e=\frac{16}{6}$$

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

$$e=\frac{8}{3}$$

Compare e and \frac{8}{3}.

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