The sum sum limits_{n=1}^{infty} frac{2n^2+3n | vedclass

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(B) We want to evaluate $S = \sum \limits_{n=1}^{\infty} \frac{2n^2+3n+4}{(2n)!}$.First,express the numerator in terms of $2n(2n-1)$:$2n^2+3n+4 = \fra

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$\frac{13e}{4}+\frac{5}{4e}-4$

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(B) We want to evaluate $S = \sum \limits_{n=1}^{\infty} \frac{2n^2+3n+4}{(2n)!}$.First,express the numerator in terms of $2n(2n-1)$:$2n^2+3n+4 = \frac{1}{2}(2n)(2n-1) + 4n + 4 = \frac{1}{2}(2n)(2n-1) + 2(2n) + 4$.Thus,$S = \sum \limits_{n=1}^{\infty} \frac{\frac{1}{2}(2n)(2n-1) + 2(2n) + 4}{(2n)!} = \frac{1}{2} \sum \limits_{n=1}^{\infty} \frac{1}{(2n-2)!} + 2 \sum \limits_{n=1}^{\infty} \frac{1}{(2n-1)!} + 4 \sum \limits_{n=1}^{\infty} \frac{1}{(2n)!}$.Recall the series expansions:$e = \sum_{k=0}^{\infty} \frac{1}{k!}$ and $e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}$.$e+e^{-1} = 2 \sum_{n=0}^{\infty} \frac{1}{(2n)!} = 2(1 + \sum_{n=1}^{\infty} \frac{1}{(2n)!}) \implies \sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{e+e^{-1}-2}{2}$.$e-e^{-1} = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} \implies \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{e-e^{-1}}{2}$.For the first term,$\sum_{n=1}^{\infty} \frac{1}{(2n-2)!} = \sum_{k=0}^{\infty} \frac{1}{(2k)!} = 1 + \sum_{n=1}^{\infty} \frac{1}{(2n)!} = 1 + \frac{e+e^{-1}-2}{2} = \frac{e+e^{-1}}{2}$.Substituting these values:$S = \frac{1}{2}(\frac{e+e^{-1}}{2}) + 2(\frac{e-e^{-1}}{2}) + 4(\frac{e+e^{-1}-2}{2})$$S = \frac{e+e^{-1}}{4} + e - e^{-1} + 2e + 2e^{-1} - 4$$S = (\frac{1}{4} + 1 + 2)e + (\frac{1}{4} - 1 + 2)e^{-1} - 4 = \frac{13}{4}e + \frac{5}{4e} - 4$.

The sum $\sum \limits_{n=1}^{\infty} \frac{2n^2+3n+4}{(2n)!}$ is equal to :

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