I'm trying to determine the convergence radius of$\sum\limits^\infty_{n=0} \frac {(-1)^n} {\sqrt{n+2} \sqrt[4]{n+4}} \cdot x^n$. The root test simplifies to $\lim\limits_{n\to\infty}\frac {1} {\sqrt[n]{\sqrt{n+2}\sqrt[4]{n+4}}} \cdot |x|$ but that doesn't lead me any further. Is there an easier way or am I missing out onto something?
How to determine the convergence radius of $\sum\limits^\infty_{n=0} \frac {(-1)^n} {\sqrt{n+2} \sqrt[4]{n+4}} \cdot x^n$
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$\begingroup$
limits
convergence
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0This way: $1$. Sorry, I couldn't resist. The coefficient of $x^n$ behaves like $n^{-3/4}$ in absolute value, hence the radius of convergence is $1$ by the very definition of it ($1/\limsup_{n\to +\infty}\sqrt[n]{|c_n|}$). – 2017-01-11
3 Answers
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$\sqrt[n]{\sqrt{n+2}}=(\sqrt[n]{n+2})^{1/2} \to 1 $ for $n \to \infty$.
$\sqrt[n]{\sqrt[4]{n+4}}=(\sqrt[n]{n+4})^{1/4} \to 1 $ for $n \to \infty$.
Your turn !
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The radius of convergence will be 1 as
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Try the ratio test instead.
$$\lim_{n\to\infty}\left|\frac{\frac {(-1)^{n+1}} {\sqrt{n+3} \sqrt[4]{n+5}} \cdot x^{n+1}}{\frac {(-1)^n} {\sqrt{n+2} \sqrt[4]{n+4}} \cdot x^n}\right| = \lim_{n\to\infty}\left|\frac{\sqrt{n+2}\sqrt[4]{n+4}}{\sqrt{n+3}\sqrt[4]{n+5}}\cdot x\right| = |x|$$
This is less than one if $|x| < 1$, and greater than one if $|x| > 1$, so the radius is $1$.
(It converges at $x = 1$ by the alternating series test and diverges at $x = -1$ by comparison to the harmonic series, but that wasn't part of the question.)