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Let's consider two sequences $(a_n), (b_n)$ in $\mathbb{R}$ such that $$\lim_{n \to +\infty} a_n, \lim_{n \to +\infty} b_n = + \infty$$

Proposition: the sequence $$x_n = \left(1 + \frac{1}{a_n}\right)^{b_n}$$ has a limit if $\lim_{n \to +\infty} \frac{b_n}{a_n}$ exists.

How can one prove it? What is this limit? Does $x_n$ have a limit if $\lim_{n \to +\infty} \frac{b_n}{a_n}$ does not exist?

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  • $\begingroup$ hint $\left(1+1/x\right)^y=e^{\log\left(1+1/x\right)y}$ $\endgroup$
    – tired
    Nov 13, 2015 at 16:38
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    $\begingroup$ This may be useful. Our power is equal to $\left(\left(1+\frac{1}{a_n}\right)^{a_n}\right)^{b_n/a_n}$.. $\endgroup$
    – André Nicolas
    Nov 13, 2015 at 16:52
  • $\begingroup$ I will post the solution soon, as soon as I find some time. $\endgroup$
    – marmistrz
    Nov 20, 2015 at 22:22

1 Answer 1

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From the definition of real power $$x_n = \left(1 + \frac{1}{a_n}\right)^{b_n} = \exp\left(b_n \ln \left(1 + \frac{1}{a_n}\right)\right) = \exp\left(\frac {b_n}{a_n} a_n \ln \left(1 + \frac{1}{a_n}\right)\right) = \exp\left(\frac {b_n}{a_n} \ln \left(1 + \frac{1}{a_n}\right)^{a_n}\right)$$ But the exponential function is continuous, so $$\lim e^{p_n} = e^r \iff \lim p_n = r$$ Notice that $$\lim \frac {b_n}{a_n} \ln \left(1 + \frac{1}{a_n}\right)^{a_n} = \lim \frac {b_n}{a_n}$$ which gives us the thesis

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