Limiting Sum maths in focus question, find t?
Hi can anyone help, I am really not too sure how they got this answer, the question is
For what values of k does the limiting sum exist for the series …? k+ k^2 + k^3....
the answer is -1<k<1
K not T!!!
- 1 year ago
S = k + k^2 + k^3 + k^4 + ...
S = k + k * (k + k^2 + k^3 + .....
S = k + k * S
S - S * k = k
S * (1 - k) = k
S = k / (1 - k)
k / (1 - k) = k + k^2 + k^3 + ....
k / (1 - k)
For infinite geometric sums, when the common ratio is between -1 and 1, the sum converges. In our sum, k is the common ratio between any 2 terms. Therefore, when -1 < k < 1, then we have a converging sum, and it converges to k / (1 - k)
- Some BodyLv 71 year ago
The series is:
k + k² + k³ + k⁴ + ...
This can be rewritten as:
k(k)⁰ + k(k)¹ + k(k)² + k(k)³ + ...
This is a geometric series, where the first term is k and the common ratio is k.
The sum of the first n terms of a geometric series is:
S = a (1 - r^n) / (1 - r)
In this case, a = k and r = k:
S = k (1 - k^n) / (1 - k)
The limit as n approaches infinity:
lim(n→∞) S =
lim(n→∞) k (1 - k^n) / (1 - k) =
k / (1 - k) lim(n→∞) (1 - k^n)
If k is a fraction, then k^n will approach 0 as n approaches infinity, so the limit will exist. Otherwise, k^n will grow to infinity, and the limit will not exist.
- D gLv 71 year ago
you can see that if the k = 1 or larger then each successive value gets larger so it will never have a limit
if the k is less than 1 then each successive value gets smaller and so if something adds to something smaller than before there is eventually a limit to the sum
since the a fraction of one times itself is always smaller this works for positive or negative