Tom K. answered • 06/03/20
Knowledgeable and Friendly Math and Statistics Tutor
One bound could be found, as this is an alternating series, by just setting the absolute value of the next term as a bound, as the alternating sum will lead to a smaller value than this. Then, setting 7/n^4 = .0001, then n^4 = 70000, or n = 10 * 7^(1/4) = 16.27. Then, as the sum after the 16th term would be less than the 17th term whose absolute value would be less than this term, you can sum the first 16 terms.
The easy way to get a limit would be to sum the first 1000 terms, know that we are within 7 * 10^-12, and compare ths sum of the first n terms. This would get you a solution of 14.
If you wanted to try to lessen the 16 number in another way, you could note that the absolute value of the ratio of successive terms goes to 1, so we can bound alternating sums by using the geometric sum found by using the ratio of the next two terms. The ratio of n^4/(n+1)^4 is approximately 1 - 4/(n + 2.5) (in fact, this is the limiting value and the actual value of the denominator is slightly greater than this), so the geometric sum is less than 1/(1 - (4/(n+2.5)-1) = 1/(2 - 4/(n+2.5)) - you get the limiting ratio by seeing that the ratio is 1 - (4n^3 + 6n^2 + 4n + 1)/(n^4+4n^3+6n^2+4n+1) and seeing that 4(1 + 3/2n)/(1+4/n) approaches 4(1+(3/2-4)/n) = 4(1-5/(2n)) is approximately 4/(1+5/(2n))
Then, we can look for the n such that 7/n^4 * 1/(2 - 4/(n+2.5)) < .0001
We see that 14 gives a value slightly above this, .000103, and 15 gives a value below this, .000078, so we use 14 terms (the sum from the 15th term on will be approximately .000078)
