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

Update:

K not T!!!

3 Answers

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  • 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)

  • 5 months 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 g
    Lv 7
    5 months 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

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