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.

 

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, 20187 yr by pavelcherepan

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

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.   

 

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, 20187 yr by StringJunky