Sum of factorials

[fsh26]

I do not know how to find the sum of factorials:

My attempt:

I do not know how correctly I chose the method, but I do not know what to do next. If anyone helps, I will be grateful.

 

[pavelcherepan]

I'm not great at maths, but it seems like you can express sum of factorials as an integral and then I guess, you can just go with sum rule in integration. Not sure what you do afterwards, though. Good luck anyway.

 

 

Edited March 24, 2018 by pavelcherepan

[fsh26]

Many thanks for this tip. I write if I can calculate this way.

[taeto]

It may no longer be useful, but I will try to add some suggestions. Sorry that I cannot make Latex work on this site.

It is clear that the sum of 1/n! for n=3,4,... is equal to e - 5/2, since we are missing just the terms for n=0,1,2 in getting the actual expansion of e as the sum of 1/n! over all natural numbers n. We just need to work out the sum of the terms of the form 1/(n! n) over n=3,4,... in your final RHS expression. Call that sum S.

Let a function f be defined by f(x) = sum{ xⁿ/(n! n) : n=1,2,... }. Then it is clear that S = f(1) - 5/4, since S is of the form of f(1), except for missing the terms with n=1 and n=2. Also f(0)=0 trivially holds.

Each term  xⁿ/(n! n) in f is the integral of t^n-1/n! dt  from t=0 to t=x. By a simple analysis argument it follows that f(x) is equal to value of the integral of (e^t - 1 - t - t²/2)/t from t=0 to t=x, which is the same as the integral of (e^t - 1)/t - 1 - t/2 from t=0 to t=x. 

Your sum S becomes the value of the integral of (e^t - 1)/t -1 - t/2 from t=0 to t=1 and then subtracting 5/4. And your sum total becomes e - 5/2 - S.   

 

[StringJunky]

59 minutes ago, taeto said:

It may no longer be useful, but I will try to add some suggestions. Sorry that I cannot make Latex work on this site.

It is clear that the sum of 1/n! for n=3,4,... is equal to e - 5/2, since we are missing just the terms for n=0,1,2 in getting the actual expansion of e as the sum of 1/n! over all natural numbers n. We just need to work out the sum of the terms of the form 1/(n! n) over n=3,4,... in your final RHS expression. Call that sum S.

Let a function f be defined by f(x) = sum{ xⁿ/(n! n) : n=1,2,... }. Then it is clear that S = f(1) - 5/4, since S is of the form of f(1), except for missing the terms with n=1 and n=2. Also f(0)=0 trivially holds.

Each term  xⁿ/(n! n) in f is the integral of t^n-1/n! dt  from t=0 to t=x. By a simple analysis argument it follows that f(x) is equal to value of the integral of (e^t - 1 - t - t²/2)/t from t=0 to t=x, which is the same as the integral of (e^t - 1)/t - 1 - t/2 from t=0 to t=x. 

Your sum S becomes the value of the integral of (e^t - 1)/t -1 - t/2 from t=0 to t=1 and then subtracting 5/4. And your sum total becomes e - 5/2 - S.   

 

If you hope to catch someone in an old post, it's best to quote that post, so that they may get a message that someone has replied. Done it for you this time.

On 24/03/2018 at 7:49 AM, fsh26 said:

Many thanks for this tip. I write if I can calculate this way.

 

Edited June 16, 2018 by StringJunky