Find the limit of expression

Question

Find the limit: $\lim \limits_{n\to\infty}(n-\sqrt[2018]{n^{2018}-n^{2017}})$

I've been trying to find solution myself and was always stuck with $\infty$. Then I've asked Wolfram Alpha about the answer and it evalated the limit to $\frac{1}{2018}$, but still I can't understand why's that so.

Answer 1

You can try the limit as $x\to0^+$ of the function, defined over (0,1), $$ f(x)=\frac{1}{x}-\sqrt[2018]{\frac{1}{x^{2018}}\frac{1}{x^{2017}}}= \frac{1-\sqrt[2018]{1-x}}{x} $$ because $$ \lim_{n\to\infty}\Bigl(n-\sqrt[2018]{n^{2018}-n^{2017}}\,\Bigr)= \lim_{n\to\infty}f\Bigl(\frac{1}{n}\Bigr) $$

On the other hand $$ \lim_{x\to0}\frac{1-\sqrt[2018]{1-x}}{x} $$ is just the derivative at $0$ of $g(x)=-\sqrt[2018]{1-x}$: $$ g'(x)=\frac{1}{2018}(1-x)^{-2017/2018} $$ so $g'(0)=1/2018$.

Answer 2

We have by binomial expansion

$$\sqrt[2018]{n^{2018}-n^{2017}}=n\left(1-\frac1n\right)^{\frac 1 {2018}}=n\left(1-\frac1{2018n}+o\left(\frac1n\right)\right)=n-\frac1{2018}+o\left(1\right)$$

then

$$n-\sqrt[2018]{n^{2018}-n^{2017}}=\frac1{2018}+o\left(1\right)$$

Answer 3

$\lim \limits_{n\to\infty}(n-\sqrt[2018]{n^{2018}-n^{2017}})$=$\lim \limits_{n\to\infty}(n-n\sqrt[2018]{1-\frac{1}{n}})$=$\lim \limits_{n\to\infty}\frac{1-(1-1/n)^{1/2018}}{1/n}$=$\lim \limits_{t\to0+}\frac{1-(1-t)^{1/2018}}{t}$ and through "Hospital's Rule this amounts to $1/2018$. General hint: Sometimes in problems like this, it is easier to first see what happens if the exponent is not so high, say $3$, and then extrapolate to a more general case

Answer 4

$$ n-(n^{2018}-n^{2017})^{1/2018}=n-(n^{2018}(1-n^{-1}))^{1/2018}=n(1-(1-n^{-1})^{1/2018}), $$ and by L'Hospital $$\frac 1 {2018} \frac{(1-n^{-1})^{1/2018-1} n^{-2}}{n^{-2}}\to \frac 1 {2018}. $$

Answer 5

let $y = \sqrt[2018]{n^{2018} - n^{2017}} = n(\sqrt[2018]{1-n^{-1}})$

$n^a - y^a = (n-y)(n^{a-1} + n^{a-2}y + \cdots + y^{a-1})\\ (n - y) = \frac {(n^a - y^a)}{n^{a-1} + n^{a-2}y + \cdots + y^{a-1}}$

Note that there are $a$ terms in the denominator.

if $a=2018$

$\frac {n^{2017}}{n^{2017} + n^{2017}(1 + n^{-1}) + \cdots + n^{2017}(1 + n^{-1})^{2017}}$

dividing numerator and denominator by $n^{2017}$ and examining the limits as $n$ goes to infinity.

$\lim_\limits{n\to \infty} \frac {1}{1 + (1 + n^{-1}) + \cdots + (1 + n^{-1})^{2017}} = \frac {1}{1+1+1+\cdots +1}$

There are 2018 terms in the denominator

$\frac {1}{2018}$

Answer 6

As others have pointed out, the binomial series method also works. $$\lim_{n\to \infty}\left[n-\left(n^{2018}-n^{2017}\right)^\frac{1}{2018}\right]=\lim_{n\to \infty}\left[n-n\left(1-\frac{1}{n}\right)^\frac{1}{2018}\right]$$

Now, using the binomial series $$n\left(1-\frac{1}{n}\right)^\frac{1}{2018}=n\left[1-\frac{1}{2018n}+\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}}{2n^2} - \frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}{\left(\frac{1}{2018}-2\right)}}{3n^3}+...\right]$$ $$=n-\frac{1}{2018}+\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}}{2n} - \frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}{\left(\frac{1}{2018}-2\right)}}{3n^2}+...$$ Therefore, $$\lim_{n\to \infty}\left[n-n\left(1-\frac{1}{n}\right)^\frac{1}{2018}\right]=\lim_{n\to \infty}\left[n-\left({n-\frac{1}{2018}+\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}}{2n} - \frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}{\left(\frac{1}{2018}-2\right)}}{3n^2}+...}\right)\right]$$ $$=\lim_{n\to \infty}\left[{\frac{1}{2018}-\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}}{2n} +\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}{\left(\frac{1}{2018}-2\right)}}{3n^2}+...}\right]$$

$$=\lim_{n\to \infty}\left(\frac{1}{2018}\right)-\lim_{n\to \infty}\left({\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}}{2n} -\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}{\left(\frac{1}{2018}-2\right)}}{3n^2}+...}\right)$$ $$=\frac{1}{2018}$$ Since $$\lim_{n\to \infty}\left({\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}}{2n} -\frac{{\frac{1}{2018}}{\left(\frac{1}{2018}-1\right)}{\left(\frac{1}{2018}-2\right)}}{3n^2}+...}\right)=0$$