Question

Calculate: `sum1/(n!+(n+1)!)` ; n from 0 to infinity?

Answer 1

`sum_(n=0)^oo 1/(n!+(n+1)!) = 1`

Explanation:

Transform the general term of the series as follows:

`1/(n!+(n+1)!) = 1/(n!+(n+1)n!)`

`1/(n!+(n+1)!) = 1/(n!(1+n+1))`

`1/(n!+(n+1)!) =1/(n!(n+2))`

`1/(n!+(n+1)!) = (n+1)/((n+2)!)`

`1/(n!+(n+1)!) = (n+2-1)/((n+2)!)`

and finally:

`(1) " " 1/(n!+(n+1)!) =1/((n+1)!) -1/((n+2)!)`

Consider now the MacLaurin series of the exponential function:

`e^x = sum_(k=0)^oo x^k/(k!)`

For `x=1` we get:

`e = sum_(k=0)^oo 1/(k!)`

extract the first term from the series:

`e = 1+ sum_(k=1)^oo 1/(k!)`

and changing the index to `n=k-1`:

`e-1 = sum_(n=0)^oo 1/((n+1)!)`

Similarly extracting the first two terms of the series we get:

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

Summing now the terms in equation `(1)` we have:

`sum_(n=0)^oo 1/(n!+(n+1)!) = sum_(n=0)^oo1/((n+1)!) - sum_(n=0)^oo1/((n+2)!)`

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

Answer 2

`1`

Explanation:

`int\ x^(n+1)/(n!) dx = 1/(n+2)x^(n+2)/(n!)` and

`sum_(n=0)^oo x^(n+1)/(n!) = x e^x` and

`int_0^x x e^x dx = 1+(x-1)e^x` and then

`S(x)=sum_(n=0)^oo1/(n+2)x^(n+2)/(n!) = 1+(x-1)e^x`

and finally

`S(1) = sum_(n=0)^oo1/((n+2)(n!)) = 1`