Factorize the denominator:
2n² + 3n + 1 = (2n + 1) (n + 1)
Expand the summand into partial fractions:
[tex]\dfrac1{2n^2+3n+1} = \dfrac2{2n+1} - \dfrac1{n+1}[/tex]
Then the sum is
[tex]\displaystyle \sum_{n=0}^\infty \frac2{(2n+1)n!} - \sum_{n=0}^\infty \frac1{(n+1)n!}[/tex]
(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,
[tex]\displaystyle e^x = \sum_{k=0}^\infty \frac{x^k}{k!}[/tex]
Integrating both sides yields
[tex]\displaystyle e^x = C + \sum_{k=0}^\infty \frac{x^{k+1}}{(k+1)k!}[/tex]
Taking x = 0 leads to C = e⁰ = 1. Move the constant to the left side and divide by x :
[tex]\displaystyle \frac{e^x - 1}x = \sum_{k=0}^\infty \frac{x^k}{(k+1)k!}[/tex]
Replacing x with x² in the series for eˣ gives
[tex]\displaystyle e^{x^2} = \sum_{k=0}^\infty \frac{x^{2k}}{k!}[/tex]
By the fundamental theorem of calculus,
[tex]\displaystyle \int_{x_0}^x e^{t^2} \, dt = C + \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)k!}[/tex]
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
[tex]\displaystyle \sum_{n=0}^\infty \frac1{(2n^2+3n+1)n!} = \boxed{2\int_0^1 e^{t^2} \, dt - (e-1)}[/tex]
and you could write the integral in terms of the so-called error function,
[tex]\mathrm{erf}(z) = \displaystyle \frac2{\sqrt\pi} \int_0^z e^{-t^2} \, dt[/tex]
or perhaps more appropriately, the imaginary error function,
[tex]\mathrm{erfi}(z) = \dfrac{\mathrm{erf}(iz)}i[/tex]
