Calculate the limit of $a_n = n(2^{1/n}-1)$

Question

On an exam we were asked to prove the sequence converges and find the value of: $$\lim_{n\to \infty}a_n = n(2^{1/n}-1)$$

On the test I tried every common convergence test we learned, to no avail.

Monotone and bounded, pinch theorem, n-th root, ratio test, and so on.

Does anyone have any idea how to solve this?

Thanks

Answer 1

Solution without L'Hopital (directly) by noticing that this is a hidden way of asking for the derivative of $f(x) = 2^x$ at $x = 0$.

$$\lim_{n \to \infty } n(2^{\frac{1}{n}} - 1) = \lim_{n \to \infty }\frac{2^{\frac{1}{n}} - 1}{\frac{1}{n}}$$

Because $\lim_{n \to \infty} \frac{1}{n} = 0$,

$$\lim_{n \to \infty }\frac{2^{\frac{1}{n}} - 1}{\frac{1}{n}} = \lim_{x \to 0} \frac{2^x - 1}{x} = \lim_{x \to 0} \frac{2^x - 2^0}{x - 0} = [\frac{d}{dx}2^x]^{x=0} = ln(2)$$

Answer 2

Note that by standard limit since $$\frac1n \to 0$$ we have that

$$n(2^{1/n}-1)=\frac{2^{1/n}-1}{\frac1n}\to \log 2$$

indeed for $x\to 0$ by l'Hopital

$$\lim_{x\to0} \frac{2^x-1}{x}=\lim_{x\to0} \frac{2^x\log 2}{1}=\log 2$$

Answer 3

More of a comment than an answer, but...

What course is this for?

If this is a calculus class where you've learned l'Hôpital or the calculus of exponentials and logarithms, then those tools will give you the answer, the details of which are seen in the other answers here.

On the other hand if this is an analysis course and you want to prove it from first principles, for example to compute the derivative of the logarithm, then that answer is entirely unacceptable, because it's circular. In that case, you can use convexity of the exponential, if you know it, as here. Or, as you mention you attempted in your question, prove that the sequence is bounded and increasing, as here. I think Bernoulli's inequality, the binomial theorem, Jensen's inequality, or AM-GM (which are all basically the same property of exponentials) may also get you there.

Answer 4

If we put $b_n=\frac {1}{n}\ln (2)$, then

$$a_n=(e^{b_n}-1)\frac {\ln (2)}{b_n} $$

now, observe that $$\lim_{n\to+\infty}b_n=0$$ and $$\lim_{x\to 0}\frac {e^x-1}{x}=1$$

thus $$\lim_{n\to+\infty}a_n=\ln (2) $$