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Approximate the sum of the series \sum_{n=1}^{\infty} ( (-1)^{n} )/( 3^{n} n! ) to four decimal...

Question:

Approximate the sum of the series

$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{3^{n}n!} $$

to four decimal places.

Alternating Series Test:

To check for convergence and divergence of the series {eq}\sum\limits_{n = 1}^\infty {{l_n}} {/eq}, the series can be represented as {eq}{l_n} = {( - 1)^{n + 1}}{k_{n\,\,\,\,\,\,\,}}{\text{or }}{l_n} = {( - 1)^n}{k_{n\,\,\,\,\,}} {/eq}, where {eq}{k_n} \geqslant 0,\forall n,\, {/eq}.

The following conditions must be satisfied for the series to converge.

(1){eq}\mathop {\lim }\limits_{n \to \infty } {k_n} = 0 {/eq}

(2){eq}\left\{ {{k_n}} \right\} {/eq} is a decreasing sequence.

Answer and Explanation:

Given that: {eq}\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{{3^n}n!}}} {/eq}

{eq}\displaystyle \eqalign{ & {\text{Let,}} ...

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