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. 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

This implies, for all \$k\$,

\$\$|S-S_k|   Olivier OloaOlivier Oloa

\$endgroup\$ Tim ThayerTim Thayer

\$endgroup\$

submit

concern feed rwandachamber.orgematics stack Exchange works finest with JavaScript allowed 