1

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?

user587389
  • 893
  • 8
  • 22
  • I made a typo, so in the expression I wrote (6 min ago), there should be parentheses. It definitely isn't convergent for even $n$ in general, take $n=2$. – PinkyWay Sep 29 '20 at 18:40
  • @Invisible So is $x_n$ convergent or divergent? – user587389 Sep 29 '20 at 18:42
  • I'm not sure if this is valid, for odd $r>1$, if we rewrite: $\lim\limits_{n\to\infty}\frac{\left(\frac2n+1\right)^r-1-\left(\left(-\frac2n+1\right)^r-1\right)}{\frac2n}\cdot\frac2nn^{r-1}=\lim\limits_{n\to\infty}\left(\frac{\left(\frac2n+1\right)^r-1}{\frac2n}+\frac{\left(-\frac2n+1\right)^r-1}{-\frac2n}\right)2n^{r-2}=4\lim\limits_{n\to\infty}rn^{r-2}$ might be divergent, where I used $\lim\limits_{x\to 0}\frac{(1+x)^a-1}x=a$. – PinkyWay Sep 29 '20 at 19:02
  • I apologize for my laziness, if you compute $x_n$ for small odd $r$ you'll see $x_n$ is divergent. – PinkyWay Sep 29 '20 at 19:23
  • Here is a list of indeterminate forms. (: – PinkyWay Sep 29 '20 at 21:31

1 Answers1

1

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.

Alann Rosas
  • 5,421
  • 1
  • 9
  • 28
  • I think the odd case could've been shortened via the standard table limit if I haven't missed something in the comment. – PinkyWay Sep 29 '20 at 19:49
  • ''So we can't just multiply $\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r$ with $n^{r-1}$" - Can you please elaborate this point? If a sequence converges to $0$ and another sequence tends to $\infty$, can't we make any definite conclusion about the convergence of their product? – user587389 Sep 29 '20 at 21:17
  • 1
    @user587389, $0\cdot\infty$ is an indeterminate form. Consider $a_n=\frac1n$ and $b_n=n$. Then $\lim\limits_{n\to\infty}\frac1n\cdot n=1$ not $0\cdot\infty$. Then, take $a_n=\frac1n$ and $b_n=n^2$ and $\lim\limits_{n\to\infty}\frac1n\cdot n^2=\lim\limits_{n\to\infty}n=\infty$, not $0\cdot\infty$. – PinkyWay Sep 29 '20 at 21:25