Find the radius and interval of convergence calculus power series
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Find the radius and interval of convergence calculus power series

[From: ] [author: ] [Date: 12-05-12] [Hit: ]
| a_{n+1}/a_n | = [|2x|^(n+1)/(n+1)!]/[|2x|^n/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.......
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!

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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.
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