Let $\{a_n\},\{b_n\}$ be the convergent sequence in $\mathbb{R}$ or $\mathbb{C}$, and $\forall n, a_n \not = 0$.

Does the sequence $\{ {a_n}^{b_n} \}$ also converge?

If it does, Prove it and if not, under what condition does it make sense?

I think it is true, and it is very important theorem in real-analysis, but any text doesn't mention it.

  • 3
    $\begingroup$ Intuitively I expect that bad things can happen if both $a_n$ and $b_n$ converge to $0$. $\endgroup$
    – celtschk
    Aug 9, 2015 at 15:17
  • $\begingroup$ For clarification: do you mean convergent sequences instead of series? $\endgroup$
    – Hirshy
    Aug 9, 2015 at 15:22
  • $\begingroup$ You're gonna need to add some restrictions. Let $\{a_n\}$ be any convergent series and $b_n = 0$ for all $n$. Then $\{a_n^{b_n}\}$ diverges. $\endgroup$ Aug 9, 2015 at 15:22
  • $\begingroup$ $a_n=0$, $b_n=0$ for all $n\in \mathbb{N}.$ $\endgroup$ Aug 9, 2015 at 15:23
  • $\begingroup$ oh, i have to edit it $\endgroup$ Aug 9, 2015 at 15:31

1 Answer 1


I think (it's not very clear) you're asking, given two convergent two convergent sequences $a_{n} \to a$ and $b_{n} \to b$, under what circumstances does $a_{n}^{b_{n}}$ tend to $a^{b}$.

Well, this is the same as asking where the function $f(x,y)=x^{y}=e^{y\log(x)}$ is continuous. But since $\exp$ is continuous everywhere, multiplication is continuous everywhere and $\log$ is continuous on $\mathbb{R}_{>0}$ (positive reals), the conditions you want are simply $a_{n}>0$ and $a_{n} \not \to 0$

Edit: It's been bugging me that this might not quite satisfy the OP, since we could have, for example, sequences like $a_{n}=-1$ and $b_{n}=1$ for all $n$, where $a_{n}^{b_{n}}$ certainly converges. Here, the ordinary definition of exponentiation (as repeated multiplication, and taking roots) disagrees with the formal definition (which requires $x>0$). So there are some weird cases, where $b_{n}$ consists entirely of fractions which have odd denominators when written in lowest terms, but we would generally leave these undefined until sequences in $\mathbb{C}$ have a rigorous footing.

There are also circumstances under which $a_{n}^{b_{n}}$ may converge to something different to $a^{b}$. We continue to not consider $a_{n}<0$, but we will consider $a_{n} \to 0$. Clearly, if $a_{n} \to 0$ and $b_{n} \not \to 0$, $a_{n}^{b_{n}} \to 0$. If we allow both sequences to tend to $0$, then it is shown in the comments that $a_{n}^{b_{n}}$ can take any finite value, or may fail to converge at all. I don't think it's likely that any further conditions will be found, as this sort of feels like "under what conditions does an infinite series converge" (perfectly well formed question, nobody has an answer).

  • $\begingroup$ There are also other examples with convergence, despite only using positive vaues. For example with $a_n=b_n=1/n$, $a_n^{b_n}$ converges to $1$. Indeed, I've not been able to find any two strictly positive series converging to $0$ where the power does not converge to $1$ (which doesn't necessarily mean they don't exist, only that you probably have to think harder to find one). $\endgroup$
    – celtschk
    Aug 9, 2015 at 16:36
  • $\begingroup$ Let $a_{n}=\frac{1}{n}$ and $b_{n}=\frac{x}{\log(n)}$ for $n>1$. Then $a_{n}^{b_{n}} \to e^{-x}$ $\endgroup$ Aug 9, 2015 at 16:44
  • $\begingroup$ In the case $a=b=0$, there is no way of setting a consistent value to $0^{0}$ in this context. $\endgroup$ Aug 9, 2015 at 16:45
  • $\begingroup$ But the question was only about the sequence being convergent, not about the value it converges to. $\endgroup$
    – celtschk
    Aug 9, 2015 at 16:51
  • $\begingroup$ But good example for another convergent value than $1$ (and of course that can easily be made into an example of non-convergence by interleaving two such sequences). $\endgroup$
    – celtschk
    Aug 9, 2015 at 16:53

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