consider the series sum from n=1 to infinity 2^nx^n / n!
find the radius of convergence R.
what is the interval of convergence (in interval notation)
walk me step by step? Thanks!
find the radius of convergence R.
what is the interval of convergence (in interval notation)
walk me step by step? Thanks!
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Note that 2^n x^n = (2x)^n
Use the Ratio Test:
| a_{n+1}/a_n | = [|2x|^(n+1)/(n+1)!]/[|2x|^n/n!] = |2x|/n
For any given x. this ratio remains less than 1 as n becomes arbitrarily large (indeed, it goes to 0). So this series converges for all real x. The interval of convergence is (-∞, ∞) and the radius of convergence R is ∞ (though I'm uncomfortable writing that, since infinity isn't a number).
The series is in fact the power series for e^(2x) - 1.
Use the Ratio Test:
| a_{n+1}/a_n | = [|2x|^(n+1)/(n+1)!]/[|2x|^n/n!] = |2x|/n
For any given x. this ratio remains less than 1 as n becomes arbitrarily large (indeed, it goes to 0). So this series converges for all real x. The interval of convergence is (-∞, ∞) and the radius of convergence R is ∞ (though I'm uncomfortable writing that, since infinity isn't a number).
The series is in fact the power series for e^(2x) - 1.