Limit of the quotient of two sequences

**Krizalid** · August 24th 2009, 07:21 PM

Let $a_n,b_n$ be two sequences verifying that $a_n=\int_{\frac1{n+1}}^{\frac1n}\arcsin(nx)\,dx$ and $b_n=\int_{\frac1{n+1}}^{\frac1n}\arctan(nx)\,dx.$ Compute $\frac{a_n}{b_n}$ as $n\longrightarrow\infty.$

Note: you can't use the MVT.

**luobo** · August 24th 2009, 07:50 PM

Originally Posted by Krizalid

Let $a_n,b_n$ be two sequences verifying that $a_n=\int_{\frac1{n+1}}^{\frac1n}\arcsin(nx)\,dx$ and $b_n=\int_{\frac1{n+1}}^{\frac1n}\arctan(nx)\,dx.$ Compute $\frac{a_n}{b_n}$ as $n\longrightarrow\infty.$

Note: you can't use the MVT.

You can get the result easily using L'Hospital's Rule (of course, do a simple transformation first), which is 2.

**Krizalid** · August 24th 2009, 07:58 PM

You said it, easily, and yes, that's the answer.

But that's quite boring!! That's why I posted the problem here! Try it without those things.

**luobo** · August 24th 2009, 08:06 PM

Originally Posted by Krizalid

You said it, easily, and yes, that's the answer.

But that's quite boring!! That's why I posted the problem here! Try it without those things.

Are you going to use something simple like the squeeze theorem or something more complicated? haha

**simplependulum** · August 24th 2009, 10:36 PM

when $n \to \infty$ ,

can we say the integral $a_n = \arcsin(n/n) \delta x$ and $b_n = \arctan(n/n) \delta x$

thus the limit is $\frac{ \arcsin(1)}{\arctan(1)} = 2$ ?

In fact , we are able to evaluate those integrals no matter what the up. and low. limit are .

**red_dog** · August 24th 2009, 10:52 PM

Let $f:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ f(x)=\arcsin(nx)$ .

f is strictly increasing. Then

$\frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arcsin\frac{n}{n+1}\le q\arcsin(nx)\leq\frac{\pi}{2}\Rightarrow$

$\Rightarrow\frac{1}{n(n+1)}\arcsin\frac{n}{n+1}\le q a_n\leq\frac{\pi}{2n(n+1)}$ (1)

Let $g:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ g(x)=\arctan(nx)$ .

g is strictly increasing. Then

$\frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arctan\frac{n}{n+1}\le q\arctan(nx)\leq\frac{\pi}{4}\Rightarrow$

$\Rightarrow\frac{1}{n(n+1)}\arctan\frac{n}{n+1}\le q b_n\leq\frac{\pi}{4n(n+1)}\Rightarrow$

$\Rightarrow\frac{4n(n+1)}{\pi}\leq\frac{1}{b_n}\le q\frac{n(n+1)}{\arctan\frac{n}{n+1}}$ (2)

Multiply inequalities (1) and (2):

$\frac{4\arcsin\frac{n}{n+1}}{\pi}\leq\frac{a_n}{b_ n}\leq\frac{\pi}{2\arctan\frac{n}{n+1}}$

Now apply limit as $n\to\infty$ and we get $\lim_{n\to\infty}\frac{a_n}{b_n}=2$

**luobo** · August 25th 2009, 04:25 AM

Originally Posted by red_dog

Let $f:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ f(x)=\arcsin(nx)$ .

f is strictly increasing. Then

$\frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arcsin\frac{n}{n+1}\le q\arcsin(nx)\leq\frac{\pi}{2}\Rightarrow$

$\Rightarrow\frac{1}{n(n+1)}\arcsin\frac{n}{n+1}\le q a_n\leq\frac{\pi}{2n(n+1)}$ (1)

Let $g:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ g(x)=\arctan(nx)$ .

g is strictly increasing. Then

$\frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arctan\frac{n}{n+1}\le q\arctan(nx)\leq\frac{\pi}{4}\Rightarrow$

$\Rightarrow\frac{1}{n(n+1)}\arctan\frac{n}{n+1}\le q b_n\leq\frac{\pi}{4n(n+1)}\Rightarrow$

$\Rightarrow\frac{4n(n+1)}{\pi}\leq\frac{1}{b_n}\le q\frac{n(n+1)}{\arctan\frac{n}{n+1}}$ (2)

Multiply inequalities (1) and (2):

$\frac{4\arcsin\frac{n}{n+1}}{\pi}\leq\frac{a_n}{b_ n}\leq\frac{\pi}{2\arctan\frac{n}{n+1}}$

Now apply limit as $n\to\infty$ and we get $\lim_{n\to\infty}\frac{a_n}{b_n}=2$

nice job! essentially the squeeze theorem.

**Krizalid** · August 25th 2009, 05:57 AM

That's nice, different solutions.

I still have a different one, I'll post it when I get back home.

**Wilmer** · August 25th 2009, 08:31 AM

I did that in my head...took 2.3333... seconds;
but I forgot how

**Krizalid** · August 25th 2009, 03:00 PM

As promised:

Put $a_n=\frac1n\int_{1-\frac1{n+1}}^1f$ and suppose that $f$ is continuous at $x=1.$ We have $n^{2}a_{n}=n\int_{1-\frac{1}{n+1}}^{1}{f}=\frac{n}{n+1}\cdot (n+1)\int_{1-\frac{1}{n+1}}^{1}{f},$ so let's prove that $(n+1)\int_{1-\frac{1}{n+1}}^{1}{f}\longrightarrow f(1)$ as $n\longrightarrow\infty.$

Note that is $(n+1)\int_{1-\frac{1}{n+1}}^{1}{f}=(n+1)\int_{1-\frac{1}{n+1}}^{1}{f(1)\,dx}+(n+1)\int_{1-\frac{1}{n+1}}^{1}{\big(f(x)-f(1)\big)\,dx},$ and besides given $\epsilon>0$ we have

$\begin{aligned}\left| (n+1)\int_{1-\frac{1}{n+1}}^{1}{f}-f(1) \right|&\le(n+1)\int_{1-\frac{1}{n+1}}^{1}{\big|f(x)-f(1)\big|\,dx} \\ & <(n+1)\int_{1-\frac{1}{n+1}}^{1}{\epsilon\,dx} \\ & =\epsilon. \quad\blacksquare\end{aligned}$

Hence $\underset{n\to \infty }{\mathop{\lim }}\,n^{2}a_{n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n+1}\cdot (n+1)\int_{1-\frac{1}{n+1}}^{1}{f}=1\cdot f(1)=f(1).$

Finally, on $a_n$ put $x\longmapsto\frac xn$ and we'll get $a_{n}=\frac{1}{n}\int_{1-\frac{1}{n+1}}^{1}{\arcsin(x)\,dx},$ in the same fashion for $b_n,$ thus $\underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{b_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n^{2}a_{n}}{n^{2}b_{n}}=\frac{\pi }{2}\cdot \frac{4}{\pi }=2.$

**luobo** · August 25th 2009, 04:12 PM

Define
$<br /> F(x)=\int_0^x f(t) \ dt<br />$

$<br /> f(1)=F'(1)=\lim_{n\to\infty} \frac{F(1)-F(\frac{n}{n+1})}{1-\frac{n}{n+1}}=\lim_{n\to\infty} (n+1)\int_{\frac{n}{n+1}}^{1}{f(x)}\;dx$ $=\lim_{n\to\infty}\frac{n+1}{n} \cdot \lim_{n\to\infty} n\int_{\frac{n}{n+1}}^{1}{f(x)}\;dx=\lim_{n\to\inf ty} n\int_{\frac{n}{n+1}}^{1}{f(x)}\;dx<br />$

**luobo** · August 25th 2009, 05:27 PM

Summary, the following methods have been used:
(1) Mean-value Theorem (not allowed here);
(2) L'Hospital's Rule;
(3) Squeeze Theorem;
(4) Definition of Limit;
(5) Definition of Derivative;
(6) .....?

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