Radius of convergence for n^n

Question

Given is a power series. We need to find the radius of convergence for this series. The series given is:

$$ \sum_{n=0}^{\infty}n^n(x-1)^n$$

To find the radius of convergence, I have first tried to substitute $y = x - 1$, since this was explained the the course notes. Then I took the limit of the absolute value of $\frac{c_k}{c_{k+1}}$ where $n \xrightarrow[]{} \infty$, where $c_k$ is $n^n$. Then I rewrote the sum as:

$$ \sum_{n=0}^{\infty}n^n(y)^n $$

Thuss the limit if I'm not mistaken is:

$$ \lim_{n \xrightarrow[]{} \infty} \mid{\frac{n^n}{(n + 1)^{n + 1}}}\mid $$

This is where I'm quite stuck. I have build up this limit, but then the course says we must use the following where

$$ \lim_{n \xrightarrow[]{} \infty} \left( \frac{n + 1}{n} \right)^n = e$$

Can anyone explain to me how this is related to each other?

Thanks in advance.

Answer 1

The Root Test on this series can help illuminate the radius of convergence. As long as the limit described by this test is greater than 1, the series diverges. Thus we have:

$$\lim_{n\to\infty} |n^ny^n|^{\frac{1}{n}} > 1$$

Simplifying we get:

$$\lim_{n\to\infty} ny$$

If y is not zero then this limit goes off to infinity for any fixed y (implying the limit is greater than 1 unless y is 0). Thus, the original series only converges if $x = 1$ .

Answer 2

$\frac {(n+1)^{n+1}} {n^{n}} =(1+\frac 1 n)^{n} (n+1) \to (e) (\infty) =\infty$.

Root test alos shows that the series diverges for all $x \neq 1$: $(n^{n}|y|^{n})^{1/n} =ny \to \infty$ for all $y \neq 0$. The radius of convergence is $0$.

In fact there is a formula for the radius of convergence $R$ of $\sum a_ny^{n}$: $R=\frac 1 {\lim \sup |a_n|^{1/n}}$.

Ref. for root test: https://en.wikipedia.org/wiki/Root_test

Answer 3

Following on your line of work we have that

$$R = \lim_{n\to \infty} \left(\frac{n+1}{n}\right)^{-n} \cdot \frac{1}{n+1}$$

If the limit on the left is finite (which it is, what does it equal?) then what does this make the radius of convergence?

Answer 4

$$R=\frac{n^n}{(n + 1)^{n + 1}}\implies \log(R)=n \log(n)-(n+1)\log(n+1)$$ Using Taylor series for large values of $n$ $$\log(R)=-\log (n)-1-\frac{1}{2 n}+O(\left(\frac{1}{n^2}\right)$$ $$R=e^{\log(R)}=\frac{1}{e n}-\frac{1}{2 e n^2}+O\left(\frac{1}{n^2}\right)$$ So $R\to 0$.

Answer 5

$a_n:=(ny)^n$; $y\not =0$, real.

Archimedean principle:

Given $M >1,$ real, there is a $n_0$ s.t.

$n_0|y| >M>1.$

For $n \ge n_0$

$n|y| \ge n_0 |y| >M$, and

$(n|y|)^n >(n_0|y|)^n >M^n>1$;

$\lim_{n.\rightarrow \infty} a_n \not =0$, $\sum a_n$ does not converge ($y \not =0$, real)