Question

approximating sum of series. pls help! T^T

Answer 1

i'm in dilenma as to whether to use 5 or 6 terms.. how to actually see how many terms to consider in order to get approximation at 4 decimal places?

Answer 2

you might wanna consider when n = odd and when n = even

Answer 3

suppose n is odd, then you have
-1/8 - 3/8^3 - 5/8^5 - 7/8^7 - ... - (2k-1) / 8^(2k-1)

Answer 4

take out negative we have:
- (1/8 + 3/8^3 + 5/8^5 + 7/8^7 + ... + (2k-1) / 8^(2k-1) )

Now let
S = (1/8 + 3/8^3 + 5/8^5 + 7/8^7 + ... + (2k-1) / 8^(2k-1) )

multiply both sides by 1/8^2 we have:
S/8^2 = (1/8^3 + 3/8^5 + 5/8^7 + 7/8^9 + ... + (2k-1) / 8^(2k+1) )

Answer 5

subtract the two equations:
S - S/8^2 = 1/8 + 2/8^3 + 2/8^5 + 2/8^7+ .. + 2/8^(2k-1) + (2k-1)/8^(2k+1)

Answer 6

now look at 2/8^3 + 2/8^5 + 2/8^7+ .. + 2/8^(2k-1), this is just a geometric series, what first term = 2/8^3 and common ratio = 1/8^2, so the infinite sum is 2/2^3 / (1 - (1/8)^2) = 1/525

so you have:
S - S/8^2 = 1/8 + 1/525 + (2k-1)/8^(2k+1)

as k appraches infinity, the last term approaches 0, so
S - S/8^2 = 1/8 + 1/525
S (1-(1/8)^2) = 65/504
S = 520/3969

but recall that it was -(S) = - (520/3969)

So the sum when n is odd is - (520/3969)

Do the same when n is even

Answer 7

when n is even, the sum is 128/3969

so the sum of the series is 128/3969 - 520/3969 = -8/81

Answer 8

yeah u got the answer! thanks! :D

Answer 9

confirmed by wolfram alpha:
https://www.wolframalpha.com/input/?i=sum_%7Bn%3D1%7D%5E%7Binf%7D+%28-1%29%5En+n%2F8%5En