How many terms of the series do we need to add in order to find the sum to the indicated...

Question:

How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

{eq}\displaystyle \sum \limits_{n\ =\ 1}^{\infty}\ (-1)^{n - 1} \frac{7}{n^4} {/eq}, {eq}\displaystyle \text{error}\ \le 0.001 {/eq}.

Alternating Series Test:

Let to check for convergence and divergence of the series {eq}\sum\limits_{n = 1}^\infty {{w_n}} {/eq} .

Then series can be represnted as {eq}{w_n} = {( - 1)^{n + 1}}{y_{n\,\,\,\,\,\,\,}}{\text{or }}{w_n} = {( - 1)^n}{y_{n\,\,\,\,\,}} {/eq}, where {eq}{y_n} \geqslant 0,\forall n,\, {/eq}.

Then for convergent of the series following condition satiesfied;

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

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

Answer and Explanation:

Given that: {eq}\displaystyle \sum\limits_{n = 1}^\infty {\frac{{7{{( - 1)}^{n - 1}}}}{{{n^4}}}} {/eq}

{eq}\displaystyle \eqalign{ &...

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