The value of the infinite series frac{1^{2}+2 | vedclass

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(C) The $n$-th term of the series is given by $t_n = \frac{\sum_{k=1}^{n+1} k^2}{(n+2)!}$.Using the formula $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{

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$\frac{5e}{6} - \frac{1}{2}$

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(C) The $n$-th term of the series is given by $t_n = \frac{\sum_{k=1}^{n+1} k^2}{(n+2)!}$.Using the formula $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$,we have $t_n = \frac{(n+1)(n+2)(2n+3)}{6(n+2)!} = \frac{(n+1)(2n+3)}{6(n+2)!} = \frac{2n^2+5n+3}{6(n+2)!}$.Alternatively,let $r = n+1$,then $t_r = \frac{\sum_{k=1}^{r+1} k^2}{(r+2)!} = \frac{(r+1)(r+2)(2r+3)}{6(r+2)!} = \frac{(r+1)(2r+3)}{6(r+2)!} = \frac{2r^2+5r+3}{6(r+2)!}$.Simplifying the expression: $t_r = \frac{2(r+2)(r+1) + r + 1 - 2}{6(r+2)!} = \frac{2}{6r!} + \frac{r+1}{6(r+2)!} = \frac{1}{3r!} + \frac{r+2-1}{6(r+2)!} = \frac{1}{3r!} + \frac{1}{6(r+1)!} - \frac{1}{6(r+2)!}$.Summing from $r=1$ to $\infty$:$S = \sum_{r=1}^{\infty} \left( \frac{2}{6r!} + \frac{3}{6(r+1)!} \right) = \frac{1}{3} \sum_{r=1}^{\infty} \frac{1}{r!} + \frac{1}{2} \sum_{r=1}^{\infty} \frac{1}{(r+1)!}$.Since $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \dots$,we have $\sum_{r=1}^{\infty} \frac{1}{r!} = e-1$ and $\sum_{r=1}^{\infty} \frac{1}{(r+1)!} = e - 1 - 1 = e-2$.$S = \frac{1}{3}(e-1) + \frac{1}{2}(e-2) = \frac{2e-2+3e-6}{6} = \frac{5e-8}{6} = \frac{5e}{6} - \frac{4}{3}$.Re-evaluating the original series sum: $t_n = \frac{2n+3}{6n!}$.$S = \sum_{n=1}^{\infty} \frac{2n+3}{6n!} = \frac{1}{6} \sum_{n=1}^{\infty} (\frac{2}{(n-1)!} + \frac{3}{n!}) = \frac{1}{6} (2e + 3(e-1)) = \frac{5e-3}{6} = \frac{5e}{6} - \frac{1}{2}$.

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