Solve 1-1/3!+1/5!-1/7!+1/9!-frac{1}{11!}=

$$1 - \frac { 1 } { 3 ! } + \frac { 1 } { 5 ! } - \frac { 1 } { 7 ! } + \frac { 1 } { 9 ! } - \frac { 1 } { 11 ! } =$$

Evaluate

$$\frac{33588829}{39916800}\approx 0.841470985$$

Solution Steps

The factorial of 3 is 6.

$$1-\frac{1}{6}+\frac{1}{5!}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

Convert 1 to fraction \frac{6}{6}.

$$\frac{6}{6}-\frac{1}{6}+\frac{1}{5!}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

Since \frac{6}{6} and \frac{1}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6-1}{6}+\frac{1}{5!}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

Subtract 1 from 6 to get 5.

$$\frac{5}{6}+\frac{1}{5!}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

The factorial of 5 is 120.

$$\frac{5}{6}+\frac{1}{120}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

Least common multiple of 6 and 120 is 120. Convert \frac{5}{6} and \frac{1}{120} to fractions with denominator 120.

$$\frac{100}{120}+\frac{1}{120}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

Since \frac{100}{120} and \frac{1}{120} have the same denominator, add them by adding their numerators.

$$\frac{100+1}{120}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

Add 100 and 1 to get 101.

$$\frac{101}{120}-\frac{1}{7!}+\frac{1}{9!}-\frac{1}{11!}$$

The factorial of 7 is 5040.

$$\frac{101}{120}-\frac{1}{5040}+\frac{1}{9!}-\frac{1}{11!}$$

Least common multiple of 120 and 5040 is 5040. Convert \frac{101}{120} and \frac{1}{5040} to fractions with denominator 5040.

$$\frac{4242}{5040}-\frac{1}{5040}+\frac{1}{9!}-\frac{1}{11!}$$

Since \frac{4242}{5040} and \frac{1}{5040} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4242-1}{5040}+\frac{1}{9!}-\frac{1}{11!}$$

Subtract 1 from 4242 to get 4241.

$$\frac{4241}{5040}+\frac{1}{9!}-\frac{1}{11!}$$

The factorial of 9 is 362880.

$$\frac{4241}{5040}+\frac{1}{362880}-\frac{1}{11!}$$

Least common multiple of 5040 and 362880 is 362880. Convert \frac{4241}{5040} and \frac{1}{362880} to fractions with denominator 362880.

$$\frac{305352}{362880}+\frac{1}{362880}-\frac{1}{11!}$$

Since \frac{305352}{362880} and \frac{1}{362880} have the same denominator, add them by adding their numerators.

$$\frac{305352+1}{362880}-\frac{1}{11!}$$

Add 305352 and 1 to get 305353.

$$\frac{305353}{362880}-\frac{1}{11!}$$

The factorial of 11 is 39916800.

$$\frac{305353}{362880}-\frac{1}{39916800}$$

Least common multiple of 362880 and 39916800 is 39916800. Convert \frac{305353}{362880} and \frac{1}{39916800} to fractions with denominator 39916800.

$$\frac{33588830}{39916800}-\frac{1}{39916800}$$

Since \frac{33588830}{39916800} and \frac{1}{39916800} have the same denominator, subtract them by subtracting their numerators.

$$\frac{33588830-1}{39916800}$$

Subtract 1 from 33588830 to get 33588829.

$$\frac{33588829}{39916800}$$

Factor

$$\frac{33588829}{2 ^ {8} \cdot 3 ^ {4} \cdot 5 ^ {2} \cdot 7 \cdot 11} = 0.841470984648068$$