Find the interval of convergence and radius of convergence of the series.
The sum from n=0 to infinity, (x-2)^n / (n^3) +1.
I had used the ratio test here but i cannot find the correct interval of convergence. I am doing something wrong. Please help.
Thanks!
The sum from n=0 to infinity, (x-2)^n / (n^3) +1.
I had used the ratio test here but i cannot find the correct interval of convergence. I am doing something wrong. Please help.
Thanks!
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Assuming that the n-th term is actually (x - 2)^n / (n^3 + 1).
Use the ratio test:
r = lim(n→∞) |[(x - 2)^(n+1) / ((n+1)^3 + 1)] / [(x - 2)^n / (n^3 + 1)]|
..= |x - 2| * lim(n→∞) (n^3 + 1) / ((n+1)^3 + 1)
..= |x - 2| * 1
..= |x - 2|.
So, this series converges when |x - 2| < 1 and diverges when |x - 2| > 1.
Check the endpoints:
x = 3 ==> Σ(n = 0 to ∞) 1/(n^3 + 1); convergent upon direct comparison with the convergent p-series Σ(n = 1 to ∞) 1/n^3.
x = 1 ==> Σ(n = 0 to ∞) (-1)^n/(n^3 + 1); absolutely convergent by the previous remarks.
So, the interval of convergence is [1, 3].
I hope this helps!
Use the ratio test:
r = lim(n→∞) |[(x - 2)^(n+1) / ((n+1)^3 + 1)] / [(x - 2)^n / (n^3 + 1)]|
..= |x - 2| * lim(n→∞) (n^3 + 1) / ((n+1)^3 + 1)
..= |x - 2| * 1
..= |x - 2|.
So, this series converges when |x - 2| < 1 and diverges when |x - 2| > 1.
Check the endpoints:
x = 3 ==> Σ(n = 0 to ∞) 1/(n^3 + 1); convergent upon direct comparison with the convergent p-series Σ(n = 1 to ∞) 1/n^3.
x = 1 ==> Σ(n = 0 to ∞) (-1)^n/(n^3 + 1); absolutely convergent by the previous remarks.
So, the interval of convergence is [1, 3].
I hope this helps!