Is the sequence always divergent Consider the sequence
$x_n = [(\frac{2}{n}+1)^r+(\frac{2}{n}-1)^r]n^{r-1}$, $n=1,2,3,..., r=1,2,3...$
Is it always true that $x_n \rightarrow\infty$ as $n\rightarrow\infty$? Or is it true only when $r$ is even number and does it converge to $0$ when $r$ is odd?
If instead the sequence was
$y_n= [1+(-1)^r]n^{r-1}$, $n=1,2,3,..., r=1,2,3,...$ then what would be the case? Would $y_n$ be divergent regardless of whether $r$ is even or odd?
 A: The first sequence $\{x_n\}$ tends to $0$ for $r=1$, and diverges to $\infty$ for all other values of $r$. The second one $\{y_n\}$ is identically $0$ for odd $r$, and diverges to $\infty$ for even $r$. I've left the solution for $\{x_n\}$ below.
If $r=1$, then $x_n=\left[\left(\frac{2}{n}+1\right)^1+\left(\frac{2}{n}-1\right)^1\right]n^{1-1}=\frac{2}{n}+1+\frac{2}{n}-1=\frac{4}{n}$, which approaches $0$. For all other values of $r$, we'll need to give separate arguments for even and odd $r$. We'll make use of the following facts:
$$\lim\limits_{n\to\infty}\left[\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r\right]=1+(-1)^r$$
$$\lim\limits_{n\to\infty}n^{r-1}=\infty$$
Case 1: $r$ is even
Here, $1+(-1)^r=1+1=2$. Since the product of a convergent sequence with nonzero limit and one that tends to $\infty$ also tends to $\infty$, we have that
$$\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}\left(\left[\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r\right]n^{r-1}\right)=\infty$$
Case 2: $r$ is odd
Note that we have already argued the case $r=1$, so $r$ is at least 3 (this will be important later).
Here, $1+(-1)^{r}=1-1=0$, so we can't just multiply $\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r$ with $n^{r-1}$ and take the limit. Instead, we'll consider the function $f(x)=\left[\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r\right]x^{r-1}$ so we can apply L'Hôpital's. This is equivalent to
$$f(x)=\frac{\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r}{x^{1-r}}$$
Both the numerator and denominator approach $0$ as $x\to\infty$, so we differentiate.
\begin{align*}
\frac{\frac{d}{dx}\left[\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r\right]}{\frac{d}{dx}x^{1-r}} &= \frac{r\left(\frac{2}{x}+1\right)^{r-1}\cdot\frac{-2}{x^2}+r\left(\frac{2}{x}-1\right)^{r-1}\cdot\frac{-2}{x^2}}{(1-r)x^{-r}}\\
&= -\frac{2rx^{-2}}{1-r}\cdot x^{r}\cdot \left[\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\right]\\
&= -\frac{2rx^{r-2}}{1-r}\left[\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\right]
\end{align*}
Notice that $\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\to 1+(-1)^{r-1}$ as $x\to\infty$. Because $r$ is odd, $r-1$ is necessarily even, so $1+(-1)^{r-1}=2$. It's clear that $x^{r-2}\to\infty$ because $r$ is at least $3$. Noting that $-\frac{2r}{1-r}>0$ because $r\geq 3$, we get
$$\lim\limits_{x\to\infty}-\frac{2rx^{r-2}}{1-r}\left[\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\right]=\infty$$
so
$$\lim\limits_{x\to\infty}f(x)=\lim\limits_{x\to\infty}\left(\left[\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r\right]x^{r-1}\right)=\infty$$
Thus,
$$\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}\left(\left[\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r\right]n^{r-1}\right)=\infty$$
This completes the solution.
