What is the radius of convergence of the series $sum_(n=0)^oo(n*(x+2)^n)/3^(n+1)$ ?

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Answer 1 · 2014-10-20T22:17:57

Let

$a_n={n(x+2)^n}/{3^{n+1}}$ . $Rightarrow a_{n+1}={(n+1)(x+2)^{n+1}}/{3^{n+2}}$

By Ratio Test,

$lim_{n to infty}|{a_{n+1}}/{a_n}|=lim_{n to infty}|{(n+1)(x+2)^{n+1}}/{3^{n+2}}cdot{3^{n+1}}/{n(x+2)^n}|$

by cancelling out common factors,

$=lim_{n to infty}|{(n+1)(x+2)}/{3n}|$

by pulling ${|x+2|}/3$ out of the limit and simplifying a bit,

$={|x+2|}/3lim_{n to infty}|1+1/n|={|x+2|}/3<1$

by multiplying by 3,

$Rightarrow |x+2|<3=R$

Hence, the radius of convergence $R$ is $3$ .

I hope that this was helpful.

What is the radius of convergence of the series sum_(n=0)^oo(n*(x+2)^n)/3^(n+1)sum_(n=0)^oo(n*(x+2)^n)/3^(n+1)?