To find the limit of a sequence

Question

Let $p>1$ and $$ x_n= n(1-2^{-\frac{1}{n}})^{p}. $$ Can someone, please help me to find $\lim_{n\to\infty}x_n$ or at least to conclude that the above limit is non-zero. I checked that $$ \lim_{n\to\infty}\frac{x_{n+1}}{x_n}=1, $$ so ratio test fails. Then, I tried Rabee's test to find, $$ \lim_{n\to\infty}n(\frac{x_{n}}{x_{n+1}}-1), $$ which reduces to $$ -\lim_{n\to\infty}\{1+\frac{(n+1)^2}{log 2} (2^{-\frac{1}{n+1}}-2^{-\frac{1}{n}})-\frac{2n+1}{n}2^{-\frac{1}{n}}\}. $$ But after the above step, I am unable to conclude the limit. Can someone please help with the same or to conclude the limit of $x_n$ is non-zero, if possible with some other method is also fine. Thanks.

Answer 1

Using the “well-known” inequality $e^x \ge 1+x$ one gets $$ 2^{-1/n} = e^{-\ln(2) /n} \ge 1 - \frac {\ln 2}{n} $$ and therefore $$ 0 \le x_n \le \frac {(\ln 2)^p}{n^{p-1}} $$ which implies that $x_n \to 0$.

Answer 2

You seem to be applying test for convergence of series. What is given is sequence, not a series.

By MVT applied to $2^{-x}$ we get $n(1-2^{-1/n})^{p} =n^{1-p}(\ln 2)^{p} 2^{-pt}$ for some $t$ between $0$ and $\frac 1 n$. Since $p>1$ we get $x_n \to 0$.

Answer 3

Define $f(x)=(1-2^{-x})^p$

Then $$x_n= \frac{f(1/n)-f(0)}{1/n} \to f'(0).$$

Can you proceed ?

Answer 4

$$\lim_{n\to\infty}\frac{\left(1-2^{-1/n}\right)^p}{\frac{1}{n}}\underbrace{=}_{l'Hopital}\lim_{n\to\infty}\frac{p\left(1-2^{-1/n}\right)^{p-1}\frac{1}{n^2}\ln2}{-\frac{1}{n^2}}=\lim_{n\to\infty} -p(\ln2)\left(1-2^{-1/n}\right)^{p-1}=0 $$