# Convergence using power-series and convergence-radius

For which values $x \in \mathbb{R}$ does the function $f(x)=k^3*3^{-k}*(x+3)^k$ converge? Use the power-series and convergence-radius approach

I have to translate the function into a power series with the format $\sum{a_k x^k}$

I seem to lack the techniques to deal with this sort of function as I am unable to get forward with this.

My Work so far: $f(x)=\frac{k^3}{3^k}*(x^k+3^k) == f(x)= k^3*3^k*x^k$

Is $k^3*3^k$ my $a_k$ in this case and have I done any mistakes?

• What is the question? To determine the limit of $k^33^{-k}(x+3)^k$ for some fixed $x$, when $k\to\infty$?
– Did
Feb 25, 2013 at 16:21
• sorry for being blurry about the question. I'm just not quite sure if im on the right track Feb 25, 2013 at 16:26

Denote $c_k= {\dfrac{k^3}{3^k}}.$ Then the radius of convergence $R$ for the series $$\sum\limits_{k=0}^{\infty}{ {\frac{k^3}{3^k}}(x+3)^k}$$ equals $$R=\lim\limits_{k\to\infty}{\left|\frac{c_k}{c_{k+1}}\right|}=\lim\limits_{k\to\infty}{{\dfrac{k^3 3^{k+1}}{3^k (k+1)^3}}}=3 \lim\limits_{k\to\infty}{{\dfrac{k^3 }{(k+1)^3}}}=3.$$ Thus general term ${ {\frac{k^3}{3^k}}(x+3)^k}\;\; \underset{k\to \infty} {\rightarrow} \;\;0$ for all $x$ such that $|x+3|<R$, i.e. $|x+3|<3$, which gives $-6<x<0.$ Therefore, functional sequence $(f_k(x))$ with terms $f_k(x)={\frac{k^3}{3^k}(x+3)^k}$ converges to $0$ for $-6<x<0.$

• how did you get the radius of convergence that quick??? :d Feb 25, 2013 at 17:07
• This seems to take the answer for granted.
– Did
Feb 25, 2013 at 17:18
• Details are in edited answer. Feb 25, 2013 at 17:26

The indication to use power series might have been misunderstood. Anyway, the basic fact leading to the solution is the following.

Fact: Let $z$ denote a complex number, $a$ a real number, and, for every $n\geqslant1$, $x_n=n^az^n$. Then $x_n\to0$ if and only if $|z|\lt1$.

You might want to prove this fact, and then to apply it to your context, choosing $a=3$ and $z=1+\frac{x}3$.