Calculate limit of $\sqrt[n]{2^n - n}$

Question

Calculate limit of $\sqrt[n]{2^n - n}$. I know that lim $\sqrt[n]{2^n - n} \le 2$, but don't know where to go from here.

Answer 1

HINT:

• $\sqrt[n]{2^n-2^{n-1}}\leq\sqrt[n]{2^n-n}\leq\sqrt[n]{2^n}$

• $\lim\limits_{n\to\infty}\sqrt[n]{2^n}=\lim\limits_{n\to\infty}2^{\frac{n}{n}}=2^1=2$

• $\lim\limits_{n\to\infty}\sqrt[n]{2^n-2^{n-1}}=\lim\limits_{n\to\infty}\sqrt[n]{2^{n-1}} =\lim\limits_{n\to\infty}2^{\frac{n-1}{n}}=2^{\lim\limits_{n\to\infty}\frac{n-1}{n}}=2^1=2$

Answer 2

Hint: $\sqrt[n]{2^n - n} = 2 (1 - \frac{n}{2^n})^{\frac{1}{n}} = 2 ((1 - \frac{n}{2^n})^{\frac{2^n}{n}} )^{\frac{1}{2^n}}$

Now... Do you know Euler?

Answer 3

The exponential function is continuous, so $$\lim_{n\rightarrow \infty}\sqrt[n]{2^n -n} = \lim_{n\rightarrow \infty} e^{\frac{\ln(2^n-n)}{n}} = e^{(\lim_{n\rightarrow \infty}\frac{\ln(2^n-n)}{n})}$$

Then you could use l'Hospital to show that $$\lim_{n\rightarrow \infty}\frac{\ln(2^n-n)}{n} = \ln(2)$$

So then you'd have $$\lim_{n\rightarrow \infty}\sqrt[n]{2^n -n} = e^{\ln(2)} = 2$$