MHB Radius of Convergence: Evaluate & Ignoring Extra Vars

[brunette15]

I am attempting to evaluate the radius of convergence for a series that goes from k=0 to infinity. The series is given by (k*x^k)/(3^k).

I have begun by using the ratio test and have gotten to the point L = (k+1)*x/3k

Now i know i can find out the radius of convergence by simply saying R = 1/L.

However, i am always unsure what to do with the extra x and k variables that i still have in L. Can they just be ignored?

 

[Nono713]

Remember that your series has $x$ as an argument, i.e. it's more a function $f$ like:
$$f(x) = \sum_{k = 0}^\infty \frac{k x^k}{3^k}$$
and the radius of convergence $R$ simply means that the series converges for all arguments $x$ such that $\lvert x \rvert < R$. So it should be clear that the radius of convergence can't depend on $k$ (which is just a dummy variable used for the sum) or $x$.

Remember that you are dealing with a power series here, which is more clearly expressed as a sum of coefficients and powers:
$$f(x) = \sum_{k = 0}^\infty \frac{k}{3^k} x^k$$
Here the power is the $x^k$ term (and is the ONLY place $x$ can appear) and the coefficient is $k / 3^k$, which must be independent of $x$. The radius of convergence of the series depends only on the coefficients, not on the $x^k$ term (which is there for all power series anyway) so the radius of convergence can't depend on $x$. Finally, the radius of convergence is defined as a limit process as $k \to \infty$, given by:
$$R = \frac{1}{\lim\sup_{k \to \infty} \sqrt[k]{\lvert a_k \rvert}}$$
where $a_k$ is the $k$th coefficient of your power series, that is, $k / 3^k$ (again, independent of $x$). So your $L$ (the limit in the denominator) is wrong, since it can't depend on $k$. Hope that helps.

 

[brunette15]

Remember that your series has $x$ as an argument, i.e. it's more a function $f$ like:
$$f(x) = \sum_{k = 0}^\infty \frac{k x^k}{3^k}$$
and the radius of convergence $R$ simply means that the series converges for all arguments $x$ such that $\lvert x \rvert \leq R$. So it should be clear that the radius of convergence can't depend on $k$ (which is just a dummy variable used for the sum) or $x$.

Remember that you are dealing with a power series here, which is more clearly expressed as a sum of coefficients and powers:
$$f(x) = \sum_{k = 0}^\infty \frac{k}{3^k} x^k$$
Here the power is the $x^k$ term (and is the ONLY place $x$ can appear) and the coefficient is $k / 3^k$, which must be independent of $x$. The radius of convergence of the series depends only on the coefficients, not on the $x^k$ term (which is there for all power series anyway) so the radius of convergence can't depend on $x$. Finally, the radius of convergence is defined as a limit process as $k \to \infty$, given by:
$$R = \frac{1}{\lim\sup_{k \to \infty} \sqrt[k]{\lvert a_k \rvert}}$$
where $a_k$ is the $k$th coefficient of your power series, that is, $k / 3^k$ (again, independent of $x$). So your $L$ (the limit in the denominator) is wrong, since it can't depend on $k$. Hope that helps.

That makes a lot more sense! Thankyou!

 

[Prove It]

Usually radii of convergence are found with the ratio test, rather than the root test. To evaluate the radius of convergence, you need to solve for x where $\displaystyle \begin{align*} \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \end{align*}$.