Infinite Series and Sigma Notation

[Guest]

I don't know if it's possible to put symbols in this, so I have to do without them.
The question says:
Solve for x if infinity symbol over sigma over k=1 x^k = 1/3
I don't know how to solve this, and when I've tried I haven't come up with the correct answer, which is 1/4.

 

[stapel]

I will guess that you mean the following:

. . . . .sum_{[k=1 to inf.]} x^k = 1/3

The first few terms of the sequence are x, x², x³, and x⁴. With x being some fixed value, this means that the sequence is geometric, with x = r, the common ratio, and x = a, the first term of the sequence.

Since this infinite geometric sequence sums to a finite value, then we can use the geometric-series summation formula:

. . . . .a/(1 - r)

For instance:

. . . . .sum_{[n=1 to inf.]} (1/2)ⁿ

...is geometric, with r = 1/2 and a = 1/2, so the sum is (1/2)/(1 - 1/2) = (1/2)/(1/2) = 1.

In your case, you have:

. . . . .x/(1 - x) = 1/3

Cross-multiply and solve for x.

Eliz.

 

[Guest]

Thank you! That's what I was looking for. I'm still not sure why x is both the common ratio and the first term, though.

 

[stapel]

I'm still not sure why x is both the common ratio and the first term, though.

Write out the first few terms. What is the first term? Divide successive terms. What is the result?

Eliz.

 

[Guest]

I see. Thanks!