Find the least number of terms for which the sum exceeds 7.998?

For the Geometric series: +2+1+1/2+... , the sum of terms will never reach 7.998.

(a million) If the complication-loose ratio is two/3 and the sum to infinity = first term / a million - complication-loose ratio then the sum to infinity is 3 ninety 9.ninety 9% of that's 2.9997 permit first term = a complication-loose ratio = r S_n = a(a million-r^n)/(a million-r) S_n = 3(a million - (2/3)^n) (2/3)^n = a million - (S_n)/3 n log(2/3) = log (a million - (S_n)/3) n = log [a million - ((S_n)/3) / log(2/3)] S_n could exceed 2.9997 n = -4 / -0.176 ... so n = 23 is the 1st term the place the series exceeds ninety 9.ninety 9% of the sum to infinity. (2) a(a million-r^sixteen)/(a million-r) = 3a(a million-r^8)/(a million-r) so r^sixteen - 3r^8 + 2 = 0 permit x = r^8 then x^2 - 3x + 2 = 0 x = a million or 2 r = a million or 2^(a million/8) r does no longer equivalent a million till we opt for a trivial answer. The ratio of the sum of the 1st 24 words to the sum of the 1st sixteen words. [a(a million-r^24)/(a million-r)] / [a(a million-r^18)/(a million-r)] = (a million - r^24) / (a million - r^sixteen) = (a million-8)/(a million-4) = 7/3

If the first term is 2, the next is 1, the next is 1/2, etc., we have:
The Nth term is 2^(2-N)
So if S is the sum of the first N terms, we have:
S = 2 + 1 + (1/2) + (1/4) + ... + 2^(2-N)
Divide that by 2:
S/2 = 1 + (1/2) + (1/4) + ... + 2^(2-N) + 2^(1-N)
Subtract that last equation from the first one:
S - (S/2) = 2 - 2^(1-N)
S/2 = 2 - 2^(1-N)
Multiply through by 2:
S = 4 - 2^(2-N)
No matter how high N is, S will never reach 4, so therefore S will never reach (or exceed) 7.998

Maybe you meant to say:
4 + 2 + 1 + (1/2) + ...
If the first term is 4, the next is 2, the next is 1, the next is 1/2, etc., we have:
The Nth term is 2^(3-N)
So if S is the sum of the first N terms, we have:
S = 4 + 2 + 1 + (1/2) + (1/4) + ... + 2^(3-N)
Divide that by 2:
S/2 = 2 + 1 + (1/2) + (1/4) + ... + 2^(3-N) + 2^(2-N)
Subtract that last equation from the first one:
S - (S/2) = 4 - 2^(2-N)
S/2 = 4 - 2^(2-N)
Multiply through by 2:
S = 8 - 2^(3-N)
So when does S exceed 7.998?
First, when does S = 7.998?
8 - 2^(3-N) = 7.998
8 - 7.998 = 2^(3-N)
0.002 = 2^(3-N)
3-N = log (base 2) of 0.002
3-N =~ -8.965784
-N =~ -8.965784 - 3
-N =~ -11.965784
N =~ 11.965784
So if N = 11, that's not good enough, but if N = 12, we have met the criteria.
So the answer is 12.

A simple Excel spreadsheet can calculate the sums to verify this:

1 4
2 6
3 7
4 7.5
5 7.75
6 7.875
7 7.9375
8 7.96875
9 7.984375
10 7.9921875
11 7.99609375
12 7.998046875

DONE!