Question

(8) $1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + ... = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} $ (b) $\frac{1}{3} - \frac{1}{18} + \frac{1}{90} - \frac{1}{324} + \frac{1}{5 \cdot 5^4} - \frac{1}{3 \cdot 5^4} + ... = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 3^n n!} $

          (8)
$1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + ... = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} $

(b)
$\frac{1}{3} - \frac{1}{18} + \frac{1}{90} - \frac{1}{324} + \frac{1}{5 \cdot 5^4} - \frac{1}{3 \cdot 5^4} + ... = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 3^n n!} $

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(8)
1 - 1 + (1)/(2) - (1)/(6) + (1)/(24) - (1)/(120) + ... = ∑n=0^∞((-1)^n)/(n!)

(b)
(1)/(3) - (1)/(18) + (1)/(90) - (1)/(324) + (1)/(5 · 5^4) - (1)/(3 · 5^4) + ... = ∑n=1^∞((-1)^n-1)/(n · 3^n n!)

Added by Austin T.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Using the table of known Maclaurin series, find the value of the each of the following infinite sums: (e) 111 11+ BT -1 (t) 11111 0 (-1-1 N1

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