How to expand in Maclaurin series this? #f(x)=x^n/(n!+(n+1)!)#

Question

509 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-23T14:05:49

#sum_(n=0)^oo x^n/((n+1)!+n!) = (1+e^x(x-1))/x^2#

Consider the MacLaurin series of the exponential function:

#e^x = sum_(n=0)^oo x^n/(n!)#

Multiply by #x#:

#xe^x = sum_(n=0)^oo x^(n+1)/(n!)#

As the series has radius of convergence #R=oo# we can integrate it term by term:

#int_0^x te^tdt = sum_(n=0)^oo int_0^xt^(n+1)/(n!)dt = sum_(n=0)^oo 1/(n!)x^(n+2)/(n+2)#

Solve now the integral by parts:

#int_0^x te^tdt = int_0^x td(e^t)#

#int_0^x te^tdt = xe^x - int_0^x e^tdt#

#int_0^x te^tdt = xe^x -e^x +1#

#int_0^x te^tdt = 1+e^x(x-1)#

Then we have:

#1+e^x(x-1) = sum_(n=0)^oo 1/(n!)x^(n+2)/(n+2)#

#1+e^x(x-1) = sum_(n=0)^oo x^(n+2)/(n!((n+1)+1))#

#1+e^x(x-1) = sum_(n=0)^oo x^(n+2)/((n+1)!+n!)#

and dividing by #x^2#:

#(1+e^x(x-1))/x^2 = sum_(n=0)^oo x^n/((n+1)!+n!)#