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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?

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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 $1 \lt \left(1+\left(\frac23\right)^n\right)^{1/n} \lt\left(1+\left(\frac23\right)^n\right) \to 1$

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  • $\begingroup$ Wow. I understand it immediately thanks. $\endgroup$
    – shyzealot
    Commented Apr 29, 2019 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
    Commented Apr 29, 2019 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 $\endgroup$
    – Ross Millikan
    Commented Apr 29, 2019 at 3:19
  • $\begingroup$ I see. Have a good day, thanks. $\endgroup$
    – shyzealot
    Commented Apr 29, 2019 at 3:24

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