Let t_{n} denote the n^{th} term of the infin | vedclass

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(B) Let the numerator of the $n^{th}$ term be $a_{n} = 1, 10, 21, 34, 49, \ldots$. This is a quadratic sequence. The first differences are $9, 11, 13,

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(B) Let the numerator of the $n^{th}$ term be $a_{n} = 1, 10, 21, 34, 49, \ldots$. This is a quadratic sequence. The first differences are $9, 11, 13, 15, \ldots$, which form an arithmetic progression. The general term $a_{n}$ is given by $a_{n} = An^{2} + Bn + C$. For $n=1, a_{1} = A + B + C = 1$. For $n=2, a_{2} = 4A + 2B + C = 10$. For $n=3, a_{3} = 9A + 3B + C = 21$. Solving these equations: $(a_{2}-a_{1}) = 3A + B = 9$ and $(a_{3}-a_{2}) = 5A + B = 11$. Subtracting gives $2A = 2 \Rightarrow A = 1$. Then $3(1) + B = 9 \Rightarrow B = 6$. Then $1 + 6 + C = 1 \Rightarrow C = -6$. So, $a_{n} = n^{2} + 6n - 6$. The $n^{th}$ term of the series is $t_{n} = \frac{n^{2} + 6n - 6}{n!}$. As $n \rightarrow \infty$, the factorial $n!$ grows much faster than the polynomial $n^{2} + 6n - 6$. Therefore, $\lim _{n \rightarrow \infty} t_{n} = \lim _{n \rightarrow \infty} \frac{n^{2} + 6n - 6}{n!} = 0$.

Let $t_{n}$ denote the $n^{th}$ term of the infinite series $\frac{1}{1 !} + \frac{10}{2 !} + \frac{21}{3 !} + \frac{34}{4 !} + \frac{49}{5 !} + \ldots$. Then $\lim _{n \rightarrow \infty} t_{n}$ is

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