Summation Problem: sum(1/[(n)(n+1)(n+1)!]), n=1 to infty

[Goistein]

A question came up a few weeks ago, it was

sum(1/[(n)(n+1)(n+1)!]) for n=1 to infinity

All I know is that it is less than e-1 but is there a way to actually find out what it comes out to? Someone said something about a Taylor expansion, but I couldn't find one.

 

[Denis]

http://www.google.ca/search?hl=en&q=Tay ... arch&meta=

 

[Goistein]

I know what the Taylor Series is, but I have no idea to what the function is. The nth derivative would be 1/[n^2(n+1)] is as far as I could go.

 

[daon]

Did you try breaking it up with partial fractions? You may get a telescoping series (have not verifyed this).

 

[Deleted member 4993]

Did you try breaking it up with partial fractions? You may get a telescoping series (have not verifyed this).

Not in this case - because it has a (n+1)! attached to it.