Question

Find the radius of converges of \[\sum_{n=1}^\infty n^nx^{n^2}\]
Please, help

Answer 1

@amistre64

Answer 2

Have you tried the ratio or root tests?

Answer 3

I did Hadamard test.

Answer 4

But not sure about \(x^{n^2}\)

Answer 5

Is it \(x^{n^2}=x^{2n}\)?

Answer 6

I had to look it up, but it strongly resembles the traditional root test. Setting up the limit, you have
\[\lim_{n\to\infty}\left|n^nx^{n^2}\right|^{1/n}=\lim_{n\to\infty} n \left|x^n\right|\]
since \(\left(n^n\right)^{1/n}=n\) and \(\left(x^{n^2}\right)^{1/n}=x^{n^2/n}=x^{n}\). The series converges for \(x\) that make the limit less than \(1\). What's the value of the limit?

Answer 7

Is it infinitive?

Answer 8

Yes, the limit is infinity. What does this tell you about the series?

Answer 9

But if I apply Hadamard with \(\lim_{n\rightarrow \infty }|\sqrt[n]{n^n}|\) . What I am not sure is about n .
On lecture notes, I have \(lim_{n\rightarrow \infty }\sqrt[n]{n^k}=lim_{n\rightarrow \infty}(\sqrt[n]{n})^k=1^k=1\)

Answer 10

So that if n =k (why not? right?) hence I still have lim =1 and then the sum converges and the radius is 1.

Answer 11

However, if counting like \(\sqrt[n]{n^n}=n\), then I got lim goes to infinitive and the sum is diverge. They conflict, right?

Answer 12

I don't think you can simply exchange \(n\) with \(k\) in the work on the board. It looks like \(k\) is fixed, but \(n\) is not (because it is constant increasing, i.e. \(n\to\infty\)).

Answer 13

So it's okay to say that \(\sqrt[n]{n^2}\to1^2\), but the reasoning does not hold if the exponent is not fixed.

Answer 14

Got you, so that, in this case, we have the series divergence and the radius is infinitive, right?

Answer 15

Right.

Answer 16

Thanks for making it clear. I do appreciate.

Answer 17

This is the way my prof solved the problem: I post to let remind later,
\[\sum_{n=1}^\infty n^n x^{n^2}=\sum_{k=1}^\infty a_kx^k\]
\[a_k=\begin{cases}n^n~~~if ~~~k=n^2\\0~~~otherwise\end{cases}\]

Answer 18

\[\sqrt[k]a_k=\begin{cases}\sqrt[n^2]n^n =n^{1/n}~~~if~~~k=n^2\\0~~~~otherwise\end{cases}\]

Answer 19

As \(n\rightarrow \infty\), \(n^{1/n}\rightarrow 1\)
hence \(lim_{k\rightarrow \infty}sup\sqrt[k]{a_k} =1\)
That shows Radius \(R=\dfrac{1}{lim_{k\rightarrow \infty}\sqrt[k]{|a_k|}}=1/1=1\)

Answer 20

Terribly sorry, I was mistaken. Apparently, the root (Hadamard) test applies for series strictly of the form \(\sum a_n x^n\), not \(\sum a_n x^{n^2}\). The latter is not a power series.