Question

`\rm \sum\limits_{ {n}=0}^(\infty) \frac{1}{(2 {n}^(2) + 3 n + 1) n!} \\`​

Answer 1

Factorize the denominator:

2n² + 3n + 1 = (2n + 1) (n + 1)

Expand the summand into partial fractions:

`\frac1{2n^2+3n+1} = \frac2{2n+1} - \frac1{n+1}`

Then the sum is

`\displaystyle \sum_(n=0)^\infty \frac2{(2n+1)n!} - \sum_(n=0)^\infty \frac1{(n+1)n!}`

(by the way, you can use \displaystyle to make things like fractions, limits, or sums/integrals render the "right" way)

Recall that for all x,

`\displaystyle e^x = \sum_(k=0)^\infty (x^k)/(k!)`

Integrating both sides yields

`\displaystyle e^x = C + \sum_(k=0)^\infty (x^(k+1))/((k+1)k!)`

Taking x = 0 leads to C = e⁰ = 1. Move the constant to the left side and divide by x :

`\displaystyle \frac{e^x - 1}x = \sum_(k=0)^\infty (x^k)/((k+1)k!)`

Replacing x with x² in the series for eˣ gives

`\displaystyle e^(x^2) = \sum_(k=0)^\infty (x^(2k))/(k!)`

By the fundamental theorem of calculus,

`\displaystyle \int_(x_0)^x e^(t^2) \, dt = C + \sum_(k=0)^\infty (x^(2k+1))/((2k+1)k!)`

where x₀ is some constant. If we choose x₀ = 0, setting x = 0 elsewhere leads to C = 0.

Putting everything together, the series has a value of

`\displaystyle \sum_(n=0)^\infty \frac1{(2n^2+3n+1)n!} = \boxed{2\int_0^1 e^(t^2) \, dt - (e-1)}`

and you could write the integral in terms of the so-called error function,

`\mathrm{erf}(z) = \displaystyle \frac2{\sqrt\pi} \int_0^z e^(-t^2) \, dt`

or perhaps more appropriately, the imaginary error function,

`\mathrm{erfi}(z) = \frac{\mathrm{erf}(iz)}i`