Using the Alternating Series Estimation Theorem we conclude that, to estimate th

Question

Using the Alternating Series Estimation Theorem we conclude that, to estimate the sum of the given series with the accuracy 0.006 . we need to find a natural number n such that bn + 1 Choices: . 0.006 . From bn + 1 = 1/5.2n + 1 + 4 Choices: . 0.006 2n + 1 Choices: . Choices: [l + (4 )(0.006 )]/[(5 )(0.006)]. [l - (4 )(0.006 )]/[(5 )(0.006)]. [(5 )(0.006 )]/[l - (4 )(0.006)]. [(5 )(0.006 )]/[l + (4 )(0.006)]. It follows that n + 1 Choices: . Choices: [ln ((5 )(0.006)) - ln (l + (4 )(0.006 ))/[ln 2]. [ln (l - (4 )(0.006 )) - ln ((5 )(0.006))/[ln 2], [ln (1 + (4 )(0.006 )) - ln ((5 )(0.006))/[ln 2], [ln ((5 )(0.006)) - ln (l - (4 )(0.006 ))/[ln 2]. The smallest integer that satisfies the above condition is n = Tries 0/30 With n = 5 we have with error less than bn + 1 = Tries 0/30

Explanation / Answer

I suppose it's the (-1)^i/(5*2^i+4) from before (not really readable here lol)
You seem thave all first correct so I ignore.

With n=5

We have sum(i=0..5) (-1)^i/(5*2^i+4) = 0.064429 (with calculator)
With error less than |b6| = 1/(5*2^6+4) = 0.003086 (since |Rn| <= |bn+1|)

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