What is the sum of k going from 1 to 3 of k!/k^k in improper fraction form?

To find the sum of ( \frac{k!}{k^k} ) for ( k ) going from 1 to 3, we need to calculate the individual values of ( \frac{k!}{k^k} ) for each integer value of ( k ) and then sum those values.

1. For ( k = 1 ):

[

\frac{1!}{1^1} = \frac{1}{1} = 1

]

2. For ( k = 2 ):

[

\frac{2!}{2^2} = \frac{2}{4} = \frac{1}{2}

]

3. For ( k = 3 ):

[

\frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9}

]

Now, we combine these fractions to find the total sum:

[

1 + \frac{1}{2} + \frac{2}{9}

]

To add these fractions, we first convert them into a common denominator. The least common multiple of the denominators ( 1,