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Functions

If a=∑_n=1^∞ ...

Question

If $a = \sum_{n = 1}^{\infty} (\frac{2 n}{(2 n - 1)!})$ and $b = \sum_{n = 1}^{\infty} (\frac{2 n}{(2 n + 1)!})$ . Find value of [a] + [3b] where [ ] denotes greatest integer function or integral value of x.

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Solution

$a = \sum_{n = 1}^{\infty} \frac{2 n}{(2 n + 1)!} = e = 2.71828 b = \sum_{n = 1}^{\infty} \frac{2 n}{(2 n + 1)!} = \frac{1}{e} = 0.36787 [a] + [3 b] = [2.71828] + 0 [3 \times 0.3678] = 2 + 1 = 3 ∴ [a] + [3 b] = 3$

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