# What is the limit of $(1+(\frac23)^n)^{1/n}$?

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?

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 $$1 \lt \left(1+\left(\frac23\right)^n\right)^{1/n} \lt\left(1+\left(\frac23\right)^n\right) \to 1$$
• You are correct it is quite different from the definition of $e$, but it is still a reasonable limit question. I would leave it Apr 29 '19 at 3:19