2
$\begingroup$

I was studying complex analysis and wanted to find the radius of convergence of the power series $$\sum_{n=1}^\infty\frac{2^n+3^n}{4^n+5^n}z^n$$ I used 'root test' and had to find the limit of the form $$\left(1+\left(\frac23\right)^n\right)^{1/n}$$ (say, it's $a_n$) It's pretty similar to the definition of $e$. And since $(\frac23)^n$ converges to $0$ more rapidly than $\frac1n$ does, I think $a_n$ must converges to $1$. Moreover, the expression like $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left[\left(1+\left(\frac23\right)^n\right)^{(\frac32)^n}\right]^{(\frac23)^n\times\frac1n}$$ is of the form $$e^0$$ and it equals to $1$. So I can conclude that the radius of convergence is $\frac53$.

But I can't give the precise reason for $a_n$ being approaching $1$. Can anybody give me the right procedure?

$\endgroup$
0
3
$\begingroup$

A simple way to show it converges to $1$ is to note that for $a \gt 1, n \gt 1, a^{1/n} \lt a$ Then you can say $1 \lt \left(1+\left(\frac23\right)^n\right)^{1/n} \lt\left(1+\left(\frac23\right)^n\right) \to 1$

$\endgroup$
4
  • $\begingroup$ Wow. I understand it immediately thanks. $\endgroup$
    – shyzealot
    Apr 29 '19 at 3:07
  • $\begingroup$ Thank you for answering me. But looking it again, I think it is too trivial question. It is completely different from the definition of e. Can I delete the whole question? $\endgroup$
    – shyzealot
    Apr 29 '19 at 3:18
  • $\begingroup$ You are correct it is quite different from the definition of $e$, but it is still a reasonable limit question. I would leave it $\endgroup$ Apr 29 '19 at 3:19
  • $\begingroup$ I see. Have a good day, thanks. $\endgroup$
    – shyzealot
    Apr 29 '19 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.