# 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

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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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 Nov 7 '11 at 17:24

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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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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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 $n$. – Peter 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. – Peter Tamaroff Feb 4 '12 at 2:34

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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