1. ## finding this limit

I have to find the limits for a whole bunch of sequences and this one has stumped me.

Find the limit as n approaches infinity of n(1 - (1 - (1/n))^(1/k)) where k is a fixed natural number.
Just for clarification, the kth root is around (1 - (1/n)).

I know that the limit is 1/k, but how do you show it algebraically?

2. $\displaystyle \displaystyle\lim_{n\to\infty}n\left(1-\sqrt[k]{1-\frac{1}{n}}\right)=\lim_{n\to\infty}\frac{1-\sqrt[k]{1-\frac{1}{n}}}{\frac{1}{n}}=$
$\displaystyle \displaystyle=\lim_{n\to\infty}\frac{\frac{1}{n}}{ \frac{1}{n}\left(1+\sqrt[k]{1-\frac{1}{n}}+\sqrt[k]{\left(1-\frac{1}{n}\right)^2}+\ldots+\sqrt[k]{\left(1-\frac{1}{n}\right)^{k-1}}\right)}=\frac{1}{\underbrace{1+1+\ldots+1}_{k} }=\frac{1}{k}$

3. Here is another way. It is messy but it works perfectly.

The idea here is to approximate $\displaystyle \left( 1 - \frac{1}{n} \right)^{1/k}$ using Taylor polynomials. We are going to approximate the function $\displaystyle (1+x)^{1/k}$ where $\displaystyle x = -\frac{1}{n}$. Note if we use the first Taylor polynomial $\displaystyle T_1(-1/n) = 1$ then this approximation leads to $\displaystyle n(1-1)=0$ which is the wrong answer because the approximation was not so good. If we use a second Taylor polynomial $\displaystyle T_2(-1/n) = 1 - \frac{1}{kn}$ then we get $\displaystyle n\left( \frac{1}{kn}\right)$ and this limit if $\displaystyle \frac{1}{k}$. Thus, it seems that second order Taylor polynomials are good enough approximations.

The above paragraph was only an idea. We need to state it formally in order to really use this. To do this we use (a weaker version) of Taylor's theorem.

Theorem: Let $\displaystyle f$ be defined on $\displaystyle (a,b)$ containing the origin. Given that $\displaystyle f$ is $\displaystyle C^{\infty}$ (i.e. it is infinitely differenciable) on $\displaystyle (a,b)$. Let $\displaystyle T_n(x)$ be the $\displaystyle n$-th Taylor polynomial. And let $\displaystyle R_{n+1}(x) = f(x) - T_n(x)$ be the remainder term. Then if $\displaystyle x\in (a,b)$ is a non-zero point then there exists $\displaystyle y$ between $\displaystyle 0$ and $\displaystyle x$ so that $\displaystyle R_{n+1}(x) = \frac{f^{(n+1)}(y)}{(n+1)!}x^{n+1}$.

So given the function $\displaystyle (1+x)^{1/k}$ on the interval $\displaystyle (-1,\infty)$ which is $\displaystyle C^{\infty}$. Notice that $\displaystyle T_1(x) = 1 - \frac{x}{k}$. The remainder term is given by $\displaystyle R_2(x) = \frac{f''(y)}{2!}x^2 = \frac{1}{2k}\cdot \left(\frac{1}{k}-1\right) (1+y)^{1/k - 2}\cdot x^2$

Here $\displaystyle x=-1/n$ so we have $\displaystyle R_2(-1/n) = \frac{1}{2k}\cdot \left(\frac{1}{k}-1\right) (1+y_n)^{1/k - 2}\cdot \frac{1}{n^2}$ where $\displaystyle y_n$ is the number gaurentted by Taylor's theorem on $\displaystyle (-1/n,0)$.

Thus,
$\displaystyle n\left[ 1 - \left( 1-\frac{1}{n} \right)^{1/k} \right] = n\left( \frac{1}{kn} + R_2(1/n)\right)$.
We have,
$\displaystyle \frac{1}{k} + nR_2(1/n)$.
It remains to show that $\displaystyle \lim \ nR_2(1/n) = 0$.
Meaning we need to show,
$\displaystyle \frac{1}{n} \cdot \frac{1}{2k} \left(\frac{1}{k}-1\right) \cdot (1+y_n)^{1/k - 2}\to 0$
This is true because $\displaystyle (1+y_n)^{1/k-2}$ is bounded on $\displaystyle (-1/n,0)$.