Radius of convergence (3x)^2 from 0 to infinity.  

The radius of convergence of the series `sum_( k = 0 )^oo (3x)^k ` is 1/3.

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I presume that your question means "radius of convergence of the power series `sum_( k = 0 )^oo ( 3x )^k .` ."

This expression is indeed a power series of the variable `x , ` which should have the form `sum_( k = 0 )^oo ( a_k x^k...

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Hello!

I presume that your question means "radius of convergence of the power series `sum_( k = 0 )^oo ( 3x )^k .` ."

This expression is indeed a power series of the variable `x , ` which should have the form `sum_( k = 0 )^oo ( a_k x^k ) . ` Our series may be written as `sum_( k = 0 )^oo ( 3^k x^k ) , ` so `a_k = 3^k . `

Each power series has its radius of convergence. It can be found as the limit of `| a_k / a_( k + 1 ) | ` provided it exists. Here the expression is simple:

`| a_k / a_( k + 1 ) | = 3^k / 3^( k + 1 ) = 1 / 3 ,`

which obviously has a limit of `1 / 3 . ` It is the radius of convergence in question.

But there is more for a power series than simply its radius of convergence. There is also the center of convergence and the behavior at the endpoints of the resulting interval. Let's find them.

The center is definitely 0 because we have powers of `x = x - 0 , ` not powers of `( x - a ) ` for some nonzero `a . ` This way the endpoints are `0 +- 1 / 3 = +- 1 / 3 . ` At the left endpoint, the series has the form `sum_( k = 0 )^oo (-1)^k , ` at the right it is `sum_( k = 0 )^oo ( 1 ) . ` Both series are divergent because their terms do not tend to zero.

Finally, we can state that the interval of convergence is exactly `( - 1/3, 1/3 ) .`

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