infinite sequences and series?

Question

a) find the partial sum S10 (sum of the first ten terms) of the series (total sum from n=1 to infinity) of 1/(n^4). Estimate the error in using S10 as an approximation to the sum of the series.

b) use the rule involving S (sum) of n and inequalities of integrals with n=10 to give an imporved estimate of the sum

c) find a value of n so that S (sum) of n is within 0.00001 of the sum.


ANY HELP ON ANY SECTION IS GREATLY APPRECIATED!

Answer 1

a) As a fraction, not reduced:

18762645622639900557312 / 17340121312772751360000

Which is a little over 1.08203658349 ...

Type zeta(4) into a Google search to get the actual value 1.0823232337111381915160036965411679027747509519187269076829762154441206161869 ...

So 1 / 1^4 + ... + 1 / 10^4 is too low by about

0.0002866502 ...

b) 1 / 12^4 + 1 / 13^4 + ... is less than the integral from 11 to positive infinity of dx / x^4 (the definite integral is 1 / (3 * 11^3)) which is itself less than 1 / 11^4 + 1 / 12^4 + 1 / 13^4 + ...

In other words,

(err - (1 / 11^4)) < (1 / (3 * 11^3)) < err = actual sum of series - S10.

Thus (1 / (3 * 11^3)) < err < ((1 / (3 * 11^3)) + 1/11^4)
0.0002504 ... < err < 0.0003187 ...
Split the difference (i.e., add the bounds and divide by two) to get just under 0.00028459 which is not a bad error estimate.

Add that to S10 to get your better estimate of the actual sum.

c) Using that same integral comparison

S(n) + (1 / (3 * (n + 1)^3)) < true sum < S(n) + (1 / (3 * (n + 1)^3)) + (1 / (n+1)^4)

Take n large enough to be certain that S(n) is within 0.00001 of the true sum,for example, 33 [from estimate] or 32 [from zeta(4)].

Dan