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I know that $$\lim_{x\rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e$$ Is there an equivalent in complex analysis for $$\lim_{|z|\rightarrow \infty}\left(1+\frac{1}{z}\right)^z=?$$

wythagoras
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Uri Goren
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    Yes, if you use the principal branch of the logarithm for the power. You have $$\biggl(1 + \frac{1}{z}\biggr)^z = \exp \biggl(z\log \biggl(1 + \frac{1}{z}\biggr)\biggr).$$ A little Taylor expansion gives the result. – Daniel Fischer Oct 06 '15 at 15:07
  • $exp(\sum{\frac{n(-1)^{n-1}}{z^{n-1}}})$, does that mean the result is $e$, regardless of the phase on $z$ ? – Uri Goren Oct 06 '15 at 16:54
  • You can use what @DanielFischer wrote plus the fact that $(\log (1+h))/h\to 1$ as $ h\to 0$ in $\mathbb C.$ – zhw. Oct 06 '15 at 17:06

1 Answers1

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For $|z|\gt1$, we can use the convergent power series $$ \log\left(1+\frac1z\right)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k\,z^k}\tag{1} $$ Then $$ \begin{align} \left|\,z\log\left(1+\frac1z\right)-1\,\right| &=\left|\,z\sum_{k=2}^\infty\frac{(-1)^{k-1}}{k\,z^k}\,\right|\\ &=\left|\,\sum_{k=1}^\infty\frac{(-1)^k}{(k+1)\,z^k}\,\right|\\ &\le\frac12\sum_{k=1}^\infty\frac1{|z|^k}\\ &=\frac1{2|z|-2}\tag{2} \end{align} $$ Therefore, $$ \lim_{|z|\to\infty}z\log\left(1+\frac1z\right)=1\tag{3} $$ and since $e^z$ is continuous at $z=1$, $$ \lim_{|z|\to\infty}\left(1+\frac1z\right)^{\large z}=e\tag{4} $$

robjohn
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