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

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

## marked as duplicate by Martin R, Dave, BallBoy, Mostafa Ayaz, hardmathFeb 16 '18 at 20:27

• Is it clearer if you rewrite as $\lim_{n\to 0^+}\frac {2^n-1}n$? – lulu Feb 15 '18 at 22:09
• – Martin R Feb 16 '18 at 8:13

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

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

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.

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

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

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