just how to estimate $sum_n=1^inftyfrac(-1)^nn^3$ with error much less than $0.01$?

In stimulate to fix the question, ns think we need to write out the terms.

So $sum_n=1^inftyfrac(-1)^nn^3=-1+frac18-frac127+frac164-...+frac(-1)^nn^3$. However I don"t watch the pattern, so exactly how are us going to calculation it?




You are watching: Approximate the sum of the given series with an error less than 0.001.

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Short answer: the fifth term is the very first term through absolute value less than $0.01$, therefore the sum can be estimated within $0.01$ of the actual worth with just the very first $4$ terms!

We recognize that this collection converges, due to the fact that it is an alternating collection whose terms are strictly decreasing and go to zero.

Let $S$ be the value of the sum. Permit $S_k$ be the $k^ extth$ partial sum, or

$$S_k=displaystylesum_n=1^k frac(-1)^nn^3$$

The fact that the series is alternative and has actually strictly diminish terms way that

$$S_k > S qquadqquad extif k equiv 0 pmod2$$and$$S_k

This implies, for all $k$,

$$|S-S_k|




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Olivier OloaOlivier Oloa

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Tim ThayerTim Thayer

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