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Suppose $a_n \to L$ in $\mathbb{R}$. Prove $2^{a_n} \to 2^L$.

I tried to start from $| 2^{a_n} - 2^L | < \epsilon$ and work my way to $|a_n - L| < \epsilon$, but logarithms are showing up and I think the problem is too simply to need to prove an inequality with a logarithm or exponential.

I cannot use the continuity of $2^n$ in this question.

Any help is appreciated.

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Let $\varepsilon>0$. Since $a_n\rightarrow L$ we can choose $N$ such that for all $n>N$ we have $|a_n-L|<\log_2(1+\frac{\varepsilon}{2^L})$

Now for this $N$ we have for all $n>N$

$$|2^{a_n}-2^L| = |2^{a_n - L + L}-2^L|=2^L |2^{a_n-L}-1|$$

Since $|a_n-L|<\log_2(1+\frac{\varepsilon}{2^L})$ we have that $|2^{a_n-L}|<1+\frac{\varepsilon}{2^L}$ which means that $1-\frac{\varepsilon}{2^L}<2^{a_n-L}<1+\frac{\varepsilon}{2^L}$ and so by the above equation $-\varepsilon <2^{a_n}-2^L <\varepsilon$, equivalently $|2^{a_n}-2^L |<\varepsilon$. This completes the proof.

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  • $\begingroup$ Why can we take the absolute value in the last step? $\endgroup$ – libby Oct 22 '18 at 13:56
  • $\begingroup$ @Libby Hmm you are right I only written one of the sides, but the other side also holds of course. Let me fix that $\endgroup$ – Yanko Oct 22 '18 at 15:13
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Let $e^b = 2$, $b>0$.

$|e^{ba_n} -e^{bL}|=$

$ e^{bL}|e^{b(a_n-L)}-1|<$

$e^{bL}(2|b(a_n-L)|)$ for

$|b(a_n-L)| <1$.

Used : $|e^x-1|<2|x$| for $|x| <1$.

1) Let $\epsilon >0$, for $\dfrac{\epsilon}{e^{bL}2b+\epsilon} >0$

there is a $n_0$ s.t. for $n \ge n_0$ :

$|a_n-L| < \dfrac{\epsilon}{e^{bL}2b+\epsilon}<1$,

2) $|e^{ba_n}-e^{bL}| <e^{bL}2b|(a_n-L)| < $

$(e^{bL}2b)\dfrac{\epsilon}{e^{bL}2b +\epsilon} <\epsilon.$

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Dividing by $2^L$ reduces this to the case when $L = 0$. So we want to show that $a_n \to 0 \implies 2^{a_n} \to 1 $.

As usual, Bernoulli to the rescue!

$(1+1/n)^n \ge 1+\frac{n}{n} = 2 $ so $2^{1/n} \le 1+\frac1{n} $.

Since $2^{1/n} > 1$, if $2^x$ is monotonic increasing, this gives us what we want. That, in turn, follows from $2^x > 0$ for all real $x$ and $2^{x+h} = 2^x 2^h > 2^x$.

Note that, more generally, if $c > 0$ then

$(1+c/n)^n \ge 1+\frac{nc}{n} =1+c $ so $(1+c)^{1/n} \le 1+\frac{c}{n} $.

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