How to Calculate the Sum of Factorial Functions Using the Floor Function?

[camilus]

Let d(n) denote the number of digits of n in its decimal representation. Evaluate the sum

\sum\limits_{n=1}^\infty \frac{1}{d(n)!}​

 

[nicktacik]

You got to show some work first, before you we can help you. Can you find any patterns in d(n) that might make this sum more manageable?

 

[CRGreathouse]

I like that problem, it's much less dumb than it appears. I'll admit that my first instinct about the closed form for the answer was off by 0.9, so check your work in case you do the same thing I do.

 

[camilus]

\sum\limits_{n=1}^\infty \frac{1}{d(n)!}={9 \over 10}(e^{10}-1)

 

[camilus]

I got the proof if anyone is interested...

 

[CRGreathouse]

Yep, that's it. I initially forgot to subtract the 1, thus my error of 0.9 mentioned above.

 

[sutupidmath]

I got the proof if anyone is interested...

yeap, i would be glad to see it!

 

[camilus]

Let d(n) denote the number of digits of n in its decimal representation. Evaluate the sum

\sum\limits_{n=1}^\infty \frac{1}{d(n)!}​

SOLUTION:

The actual summation of \frac{1}{d(n)!} looks like

\sum\limits_{n=1}^\infty \frac{1}{d(n)!} = \frac{1}{1!} + \cdots \frac{1}{2!} + \cdots \frac{1}{3!} + \cdots​

This can be analytically simplified to

\sum\limits_{n=1}^\infty \frac{1}{d(n)!} = 9(\frac{1}{1!}) + 90(\frac{1}{2!}) + 900(\frac{1}{3!}) + 9000(\frac{1}{4!}) + \cdots​

after collecting terms and simplyfing to a summation, the result is

\sum\limits_{n=1}^\infty \frac{1}{d(n)!} = 9(\frac{1}{1!}) + 90(\frac{1}{2!}) + 900(\frac{1}{3!}) + 9000(\frac{1}{4!}) + \cdots = 9\sum\limits_{n=0}^\infty \frac{10^n}{(n+1)!}​

Adding all the terms would give us \sum\limits_{n=1}^\infty \frac{1}{d(n)!} = 9\sum\limits_{n=0}^\infty \frac{10^n}{(n+1)!} = 19822.91922

My intuituion tells me that there should be a more simple representation for \sum\limits_{n=0}^\infty \frac{10^n}{(n+1)!}.

Now by definition

e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots​

dividing by x gives us

{e^x\over x} = {1 \over x}\sum_{n = 0}^{\infty} {x^n \over n!} = {1 \over x}[1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots] = 1 + {1 \over x} + {x \over 2!} + {x^2 \over 3!} + {x^3 \over 4!} + \cdots​

Now the series (1 + {1 \over x} + {x \over 2!} + {x^2 \over 3!} + {x^3 \over 4!} + \cdots) can be rewritten as \sum_{n = 0}^{\infty} {x^n \over (n+1)!} + {1 \over x}.

So {e^x\over x} = \sum_{n = 0}^{\infty} {x^n \over (n+1)!} + {1 \over x}, and solving for \sum_{n = 0}^{\infty} {x^n \over (n+1)!} gives us

\sum_{n = 0}^{\infty} {x^n \over (n+1)!} = {e^x \over x} - {1 \over x} = {e^x - 1 \over x}​

Substituting gives us

\sum\limits_{n=1}^\infty \frac{1}{d(n)!} = 9\sum\limits_{n=0}^\infty \frac{10^n}{(n+1)!} = 9[{e^{10} -1 \over 10}] = {9 \over 10}(e^{10} - 1) which indeed equals 19822.91922.

So in conclusion:

\huge \sum\limits_{n=1}^\infty \frac{1}{d(n)!} = {9 \over 10}(e^{10} - 1)​

 

[Gib Z]

I was doing stupid things :(
I wrote d(n) as the floor function of log_{10} n and tried some magic with that >.< I thought i had a good idea :(

 

[camilus]

the floor function is used to define this function.

d(n) = floor({ln (n) \over ln (10)}) + 1