# Finding $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ without L'hopital

I found the limit $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ by first defining $f(x)=\sqrt[x]{b^{2^{-x}}-1}$ above $R$ and then finding the limit of $ln(f)$ (to cancel the nth root). This worked (the result is $1/2$), but I ended up having to find the derivative of rather complex functions when I used L'hopital (twice). My worry is that if I have to solve something like this in a test I'll easily make a technical error. I was wondering if there is a simpler way to find this limit?

I know most basic techniques of finding limits in $R$ and a bit (Stoltz, Cantor's lemma, ...) about finding limits of sequences.

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I don't have the time to see if the argument goes anywhere, but you can use the fact that $a^n-b^n=(a-b)(a^{n-1}b+a^{n-2}b^2+\cdots+ab^{n-1})$, letting $a=\sqrt[n]{b^{2^{-n}}-1}$ and $b=\frac{1}{\sqrt[n]{2^n}}$. – user5137 Nov 7 '11 at 16:50
IMHO - l`Hopital is one of the simplest methods. It saves you proofs of the form "for each epsilon there exists N such that for every n>N ...". Those usually involve more complex technics – valdo Nov 7 '11 at 16:55
+1 for explaining your reasoning. We've had many questions recently that ask, inexplicably, for help with "(doing something or other) without using (simple standard first-choice technique)". – Henning Makholm Nov 7 '11 at 17:02
Tests seldom have questions that are technically very demanding. Finding a clever solution ordinarily takes more time than applying standard techniques in more or less standard ways. The problem with a truly good solution like the one by robjohn is that you may be downgraded for not filling in details. The suggestion by Eric Naslund helps. Let $y=2^{-x}$ and you are looking at the limit as $y$ approaches $0$ of $(-\ln 2)\frac{\ln(b^y-1)}{\ln y}$, still two applications of L'Hospital's Rule, but less risky differentiation. – André Nicolas Nov 7 '11 at 17:07

Use $\left(b^{2^{-n}}-1\right) 2^n \sum_{k=0}^{2^n-1} b^{k 2^{-n}} 2^{-n} = b - 1$. Notice that, by the definition of Riemann integral, $\lim_{n \to \infty} \sum_{k=0}^{2^n-1} b^{k 2^{-n}} 2^{-n} = \int_0^1 b^x \mathrm{d} x = \frac{b-1}{\log b}$.

Hence the result: $$\lim_{n \to \infty} \sqrt[n]{b^{2^{-n}}-1} = \frac{1}{2} \lim_{n \to \infty} \sqrt[n]{ \log b } = \frac{1}{2}$$

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Interesting. I had never thought to look at it this way. (+1) – robjohn Nov 7 '11 at 17:06
+1, Very clever. – Eric Naslund Nov 7 '11 at 17:24

For $x$ near $0$, the Taylor series for $e^x-1$ gives $$e^x-1=x+O(x^2)\tag{1}$$ so \begin{align} b^{2^{-n}}-1 &=e^{\log(b)\;2^{-n}}-1\\ &=\log(b)\;2^{-n}+O(4^{-n})\\ &=\log(b)\;2^{-n}(1+O(2^{-n}))\tag{2} \end{align} It is fairly easy to show that the limit of $n^{th}$ root of $(2)$ is $2^{-1}=\frac{1}{2}$

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Taking the logarithm is a very good way to start. Then we are looking at $$\lim_{n\rightarrow \infty } \frac{1}{n}\log\left(b^{2^{-n}}-1\right).$$ Do a variable change, and let $x=2^{-n}$ so that this is $$\lim_{x\rightarrow 0} -\frac{\log 2}{\log x} \log\left( b^x -1\right).$$ As $b^x=e^{x\log b} =1 +x\log b+O(x^2)$, we see that this is

$$\lim_{x\rightarrow 0} -\frac{\log 2}{\log x}(\log(x\log b)+\log(1+O(x)))=\lim_{x\rightarrow 0}-\log 2+O\left(\frac{1}{\log x}+x\right)=-\log 2.$$

Hence the original limit is $e^{-\log 2}=\frac{1}{2}$.

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By definition you have that if $b>1$ and $k= b^n$ then

$$\log x = \mathop {\lim }\limits_{n \to \infty } k\left( {{x^{1/k}} - 1} \right)$$

Thus in your case let's put

$$\log b = \mathop {\lim }\limits_{n \to \infty } {2^n}\left( {{b^{1/{2^n}}} - 1} \right)$$

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\root n \of {{2^n}\left( {{b^{1/{2^n}}} - 1} \right)} }}{{\root n \of {{2^n}} }} = \mathop {\lim }\limits_{n \to \infty } \root n \of {\left( {{b^{1/{2^n}}} - 1} \right)} = \mathop {\lim }\limits_{n \to \infty } \frac{{\root n \of {\log b} }}{2} = \frac{1}{2}$$

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Did you intend to have $\lim\limits_{k\to\infty}$ in the first limit? – robjohn Feb 4 '12 at 0:21
Well, you can see it is the same for both. If $n \to \infty$ then $k \to \infty$ and vice versa. I kept $n$ since your limit has $n$. – Pedro Tamaroff Feb 4 '12 at 1:35
If $b<1$, then as $n\to\infty$, $k\to0$, in which case $\displaystyle\lim_{n\to\infty}k\left(x^{1/k}-1\right)=\left\{\begin{array}{}\in‌​fty&\text{if }x>1\\0&\text{if }x\le1\end{array}\right.$ – robjohn Feb 4 '12 at 2:01
rob, indeed I meant to put $b>1$. I'll edit now. – Pedro Tamaroff Feb 4 '12 at 2:34
+1 nice solution, but I've never seen that definition before. By log do you mean the natural logarithm or log base 10? And did you just derive that definition and I'm just not seeing how? – Ovi Jun 24 '13 at 17:42

If I just had to come up with the answer (no showing of work required), I'd reason as follows:

• When $n$ is big, $b^{2^{-n}}$ is near 1. How close is it to 1? (We need to know so we can subtract $1$ from it.)
• Actually taking logs, $\log b^{2^{-n}} = 2^{-n} \log b$. Since $\log x \approx x-1$ for $x$ near $1$, we have $b^{2^{-n}}-1 \approx 2^{-n} \log b$.
• So the thing you're taking the limit of is $(2^{-n} \log b)^{1/n} = (1/2) (\log b)^{1/n}$; as $n \to \infty$ this approaches $1/2$.

(Writing this out more explicitly is pretty close to Peter's method.)

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If $(c_n)$ is a sequence of positive numbers such that $\lim\limits_{n\to\infty}\frac{c_{n+1}}{c_n}$ exists, then $\lim\limits_{n\to\infty}\sqrt[n]{c_n}=\lim\limits_{n\to\infty}\frac{c_{n+1}}{c_n}$. See for example Theorem 3.37 of Rudin's Principles of mathematical analysis, 3rd Ed.

If we let $c_n=b^{2^{-n}}-1$, then $$\frac{c_{n+1}}{c_n}=\frac{\sqrt{b^{2^{-n}}}-1}{b^{2^{-n}}-1}=\frac{1}{\sqrt{b^{2^{-n}}}+1}\to\frac{1}{2}.$$

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Here is a different proof:

(I am assuming $b \gt 1$).

Consider $$a_n = b^{1/2^n} + 1$$

It is easy to see that $a_n \to 2$ as $n \to \infty$ (and so $\log a_n \to \log 2$).

By using the fact that $S_n = \frac{1}{n} \sum_{k=1}^{n} s_k$ converges to the same limit as $s_n$, we have that, by considering $\log a_n$, that

$$c_n = \sqrt[n]{a_1a_2 \dots a_n} \to 2$$

Now we can see that

$$(b^{1/2^n}-1)a_n a_{n-1} \dots a_1 = b-1$$

using the identity

$$(x-1)(x+1)(x^2+1)(x^{4} + 1) \dots (x^{2^k} + 1) = x^{2^{k+1}} -1$$

We thus have

$$\sqrt[n]{b^{2^{-n}} -1} = \frac{\sqrt[n]{b-1}}{c_n}$$

Thus we see that $$\sqrt[n]{b^{2^{-n}} -1} \to \frac{1}{2}$$

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