Use alternating series to determine the smallest value that n must be to estimate this integral using the series found in problem 20 with an error of no more than 0.00001. for # 20 I found 3/2 sum...

Use alternating series to determine the smallest value that n must be to estimate this integral using the series found in problem 20 with an error of no more than 0.00001.

for # 20 I found 3/2 sum n=0 to infinity (-1)^n/((2^n( 3n+2))

Asked on by gulfem45

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tiburtius | High School Teacher | (Level 2) Educator

Posted on

If you want your error to be less than 0.00001, that means you need to solve the follow ind inequality.

`|(-1)^n/(2^n(3n+2))|leq0.00001`

Since `n` is nonnegative and `(-1)^n` does not affect the absolute value you have

`1/(2^n(3n+2))leq0.00001`  

Now we take reciprocal value of whole inequality to get

`2^n(3n+2)geq100000`

To solve this you must solve this equation.

`2^n(3n+2)=100000`

Unfortunately this equation cannot be solved exactly thus we must use some numerical method. Since this is differentiable function it would be good to use Newton method. So after using some numerical method (I won't do that here because it's quite a long process you should get something like this

`n geq 11.4283`

Which means that `n` must be at least 12 because `n` is a nonnegative integer.  

Hence your solution is `n=12.`

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