Finding the Smallest N for a Precise Partial Sum in a Series

[mattmannmf]

determine the smallest positive integer N such that the partial sum is within 10^-10 of the series sum.

{Sigma} 2(-1/4)^(n-1) < 10^-10

ok divide the 2 out

(-1/4)^(n-1) < 5^-10


I know i have to take the natural log like so:
(n-1) ln|-1/4| < nl|5^-10|

But there's the negative 1/4, how do i get it to be positive? i get the right answer when its positive but I am not sure how to do it.

 

[Dick]

It's an alternating series sum{a_n}. You have an error estimate for the partial sum up to n in terms of |a_(n+1)|. Note the absolute value.

 

[HallsofIvy]

determine the smallest positive integer N such that the partial sum is within 10^-10 of the series sum.

{Sigma} 2(-1/4)^(n-1) < 10^-10

Is \sum 2 (-1/4)^{n-1} the sum? You have stated this wrong- you don't want the partial sum, which is what you have here, to be less than 10^{-10}, you want its difference from the actual sum to be less than 10^{-10} but have then done that.

One way to do that is to find the actual sum. That is not difficult because this is a "geometric" series and there is a formula for the sum.

But, better, as Dick says, because this is alternating + and -, the "error" is always less than the next term. You just want to find n so that the absolute value of the next term is less than 10^{-10}

ok divide the 2 out

(-1/4)^(n-1) < 5^-10


I know i have to take the natural log like so:
(n-1) ln|-1/4| < nl|5^-10|

But there's the negative 1/4, how do i get it to be positive? i get the right answer when its positive but I am not sure how to do it.

\left|\left(-\frac{1}{4}\right)^{n-1}\right|= \left(\frac{1}{4}\right)^{n-1}&lt; 5^{-10}

 

[mattmannmf]

so all i have to do is take the absolute value, THEN take the natural log? and solve from there pretty much

 

[Dick]

so all i have to do is take the absolute value, THEN take the natural log? and solve from there pretty much

Pretty much, yes. Provided you understand why that works.