Find the interval of convergence of $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$

Question

How to find the interval of convergence of the following series: $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$

I have no idea what to proceed. Any help? Thanks!

Answer 1

Hint: Apply the root test to the series. If we let $a_n$ be the sequence starting $x,\frac{1}2x^2,3x^3,\ldots$, then we wish to calculate $\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}$, knowing that if this is less than $1$, the series converges, and if it is greater than one, it diverges.

Notice that the coefficients of odd powers are always the greatest so the limit supremum equals $$\lim_{n\rightarrow\infty}\sqrt[2n+1]{|a_{2n+1|}}=\lim_{n\rightarrow\infty}\sqrt[2n+1]{(2n+1)|x|^{2n+1}}=\lim_{n\rightarrow\infty}\sqrt[2n+1]{2n+1}\cdot |x|=|x|.$$ From here, you can easily find the radius of convergence, leaving just two easy boundary cases to resolve.

Answer 2

I would recommend splitting the $x^k*k$ and the $\frac{x^k}{k}$ terms and seeing where the two intervals intersect.

$(∑(2k+1)x^{2k+1}) + (∑\frac{x^{2k}}{2k})$

Answer 3

There's a formula (Cauchy) for computing the radius of convergence of arbitrary series:

$$\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{k\to\infty}\sqrt[2k]{2k}=\lim_{n\to\infty}\sqrt[n]n=1$$ It's rather clear we are only interested in odd terms because of $\limsup$ and $\sqrt[n]n\to1$ is well known, therefore the radius is $1/1=1$.