Convergence of a geometric sequence

Calculus Level 2

Consider the sequence $\left\{a_n\right\}$ where $a_n=\left(\frac{2x-1}{5}\right)^n.$ What is the sum of all integers $x$ for which this sequence converges?

The answer is 5.

3 solutions

Vipin Kumar
May 2, 2014

The sequence $r^n$ converges only if $-1 < r \leq 1$ . So we set up $-1 < \frac{2x - 1}{5} \leq 1$ . Solving this inequality gives us $-2 < x \leq 3$ . The only integers satisfying this inequality are -1, 0, 1, 2, and 3. Their sum is 5.

I don't think 1 works. If r=1 the seris will be a constant and the sum will never be converged.

yubo liu - 5 years, 3 months ago

I agree, there should be a strict inequality on the right side...the solution should be -1+0+1+2=2

james brady - 5 years, 3 months ago

I agree with you, but you are not in agreement with Yubo Liu.

Matthew Li - 5 years, 1 month ago

Agree. #VoteFor2

Nanda Rahsyad - 5 years ago

r=1 does work, because this is a sequence, not a sum. If it were a sum, you would need a strict inequality r<1. For a sequence, r=1 converges to 1.

Steve McMath - 4 years, 6 months ago

I agree with you Steve

Praveen Kumar - 2 years, 10 months ago

Its wrong what you've written

Adarsh Gkr - 7 years ago

At x=1 , series fails to remain convergent (as it becomes a constant) and hence , it should not be considered..

Rishabh Shekhar - 4 years, 1 month ago

The question refers to when the sequence converges, not when the series converges. When x=1, the sequence is 1, 1, 1, 1, etc., so the sequence converges to 1 even though it's series would diverge.

Noam Amozeg - 2 years, 2 months ago

What does it mean for a sequence to converge..? Please explain.

Puneet Pinku - 4 years ago

of course the sum doesn't converge with x=3 because the sequence then is 1^n and its sum is obviously infinite

Nik Gibson - 2 years, 10 months ago

Ohh I tottally forgot about (-1)^n, thats why the answer wasn't 3... how could I? Beautiful problem

Ezequiel Ghiena - 1 year, 8 months ago

the integers for x so that the sum converges is -3, -2, -1, 0, 1, 2 so shouldn't it be -3?

James Watson - 1 year ago

If that is the case what you've written, even -1 is an integer

Adarsh Gkr - 7 years ago

if you take -1 then limit does not exists ,so it can not be convergent!!

Kukadiya Maheshkumar - 5 years, 7 months ago

Exactly correct solution, i have done it by same way!!

Kukadiya Maheshkumar - 5 years, 7 months ago

it doesnt converge for r=1. answer should be 2.

Aishit Dharwal - 5 years ago

Question says:

sum of all the possible positive integers

In that case, 1 + 2 + 3 = 6?

Jeganathan Sriskandarajah - 5 years, 6 months ago

It doesn't specify as positive

Jeremiah Masvero - 5 years, 3 months ago

Sourayan Banerjee
May 19, 2014

This problem is easily solvable by the root test method. given series is ((2x-1)/5)^n. applying root test((2x-1)/5)^n^1/n=(2x-1)/5 To be convergent(2x-1)/5<1, so x<3. Then only two +ve integers are 1 and 2 so ans is 3.

John Dore
May 25, 2014

(n+1) term must be <1 so x=1 and x=2 are acceptable thus 1+2=3

Convergence of a geometric sequence

The answer is 5.

3 solutions

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