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For which values of $z$ does $\sum_{n=1}^{\infty} 2^{n}n^{n}z^{n} $ converge?

I know the first step is to perform a ratio test to find the radius of convergence, but I'm having trouble choosing an $a_{k}$ to find the ratio between $a_{k}$ and $a_{k+1}$ .

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4 Answers 4

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If ratio test is the first step, have a look at the zeroeth step: Does the summand tend to $0$ at all? If $z\ne0$ then for $n$ large enough $|nz|>1$ and hence $|2^nn^nz^n|>2^n$.

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  • $\begingroup$ I took the liberty of correcting two typos. I hope you don't mind. +1 for the answer. $\endgroup$
    – Mark Viola
    Nov 13, 2015 at 23:05
  • $\begingroup$ Look up Hadamard Radius Formula. $\endgroup$ Nov 14, 2015 at 1:38
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You can use the ratio test with $a_n=2^n n^n$.

However to get a better understanding of power series, I would suggest to go to the basics. If $\sum_{n=1}^{\infty} 2^{n}n^{n}z^{n} $ converges, then the term of the series has to be bounded. Which means that you can find $M >0$ such that $$\vert 2^{n}n^{n}z^{n} \vert < M$$ for all $n \in \mathbb N$, or $\vert z \vert < \frac{M^{1/n}}{2n}$. As the RHS tends to $0$, the radius of convergence is equal to $0$.

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The root test gives $\sqrt[n]{|2^{n}n^{n}z^{n}|}=|2nz| \to \infty$ and so the series diverges, unless $z=0$.

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Look at the term For $z$ such that $|z|<1,$ $$|2^n n^nz^n|=|(2n)^nz^n|\ge(2n)^n\frac 1 {n^n}=2^n>1\text{ for }n>\frac 1{|z|}$$

For $z$ such that $|z|\geq 1$, $$|2^n n^nz^n|\ge(2n)^n>1$$

And for $z=0$, the series is just zero.

Hence, the series is converges only at $z=0$.

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  • $\begingroup$ This isn't a proof per se - for $z=\frac14$, for instance, the individual terms aren't greater than $n^n$. Your result is correct but you haven't really shown it. $\endgroup$ Nov 14, 2015 at 0:19
  • $\begingroup$ thanks for the comment..I have corrected the mistake..@StevenStadnicki $\endgroup$
    – David
    Nov 14, 2015 at 0:41

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