I have a simple geometric sum:

Sum[a^(j + 1), {j, 1, k}]

which evaluates to

(a^2*(-1 + a^k))/(-1 + a)

a has a given range of -2/3<a<=1. I want to plot the limit of this function as k->Infinity for the given range of a. (Where and if such limits exist.)

Is this possible? If so, how?

(ListPlot is fine, since k is an integer.)

  • $\begingroup$ Isn't it just Plot[a^2/(1 - a), {a, -2/3, 1}] as a^k->0 for k->Infinity ? $\endgroup$ – Vitaliy Kaurov Sep 14 '18 at 9:41
  • $\begingroup$ Hm. I guess so. Silly me! OK, I'll try to work with that. Want to promote it to 'answered'? $\endgroup$ – Richard Burke-Ward Sep 14 '18 at 9:47

For the values of your parameters, -2/3<a<=1 and k > 0 && k ∈ Integers, the limit is rather simple as a^k->0 for k->Infinity:

Plot[a^2/(1 - a), {a, -2/3, 1}]

enter image description here

  • $\begingroup$ Is it really still true if $a=1$? $\endgroup$ – Johu Sep 14 '18 at 12:02
  • $\begingroup$ Limit[(a^2*(-1 + a^k))/(-1 + a) /. {a -> 1}, {k -> Infinity}] is Indeterminate $\endgroup$ – Johu Sep 14 '18 at 12:05
  • $\begingroup$ @Johu you have singularity at $a=1$ as divide by zero $\endgroup$ – Vitaliy Kaurov Sep 14 '18 at 13:09
  • $\begingroup$ This is what I thought. Yet you have $a\le 1$ in your answer, which is strictly speaking not correct. $\endgroup$ – Johu Sep 14 '18 at 13:30
  • $\begingroup$ @Johu the limit is correct - I mean the limit I described in my post. Smallness of $a^k$ is independent of singularity in the denominator. $\endgroup$ – Vitaliy Kaurov Sep 14 '18 at 13:34

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