Infinite Sequences and Series

**DemonGal711** · November 13th 2008, 09:41 PM

I have to determine whether the sequence converges or diverges and, if it does converge, find the limit. I'm confused about these two.

arctan(2x)

The limit of arctan(x) as x goes to infinity is pi/2. So, doesn't arctan(2x) do the same, just the behavior around the origin is altered? Can I just put that the limit of arctan(2x) is pi/2 even if I really didn't do anything to prove it?

ln x
ln 2x

Both go to infinity, the denominator goes just a bit quicker. So, I do l'hopital rule and I get

1/x
2/2x

1/x * 2x/2

Which reduces to 1 if I'm not mistaken. So, I get the obvious limit.

lim 1 = 1

which means my original problem's limit is 1. Is that right? It just seemed to easy to me.

**mr fantastic** · November 13th 2008, 09:49 PM

Originally Posted by DemonGal711

I have to determine whether the sequence converges or diverges and, if it does converge, find the limit. I'm confused about these two.

arctan(2x)

The limit of arctan(x) as x goes to infinity is pi/2. So, doesn't arctan(2x) do the same, just the behavior around the origin is altered? Can I just put that the limit of arctan(2x) is pi/2 even if I really didn't do anything to prove it?

[snip]

This is what you should do:

Substitute $2x = t$ :

$\lim_{x \rightarrow +\infty} \tan^{-1} (2x) = \lim_{t \rightarrow +\infty} \tan^{-1} (t) = \frac{\pi}{2}$ .

**mr fantastic** · November 13th 2008, 09:53 PM

Originally Posted by DemonGal711

[snip]
ln x
ln 2x

Both go to infinity, the denominator goes just a bit quicker. So, I do l'hopital rule and I get

1/x
2/2x

1/x * 2x/2

Which reduces to 1 if I'm not mistaken. So, I get the obvious limit.

lim 1 = 1

which means my original problem's limit is 1. Is that right? It just seemed to easy to me.

That's fine. Alternatively:

$\lim_{x \rightarrow +\infty} \frac{\ln (x)}{\ln (2x)} = \lim_{x \rightarrow +\infty} \frac{\ln (x)}{\ln (2) + \ln (x)}$

$= \lim_{x \rightarrow +\infty} \frac{1}{\frac{\ln (2)}{\ln (x)} + 1} = \frac{1}{0 + 1} = 1$ .

**DemonGal711** · November 13th 2008, 10:01 PM

Thanks. Glad to know I wasn't completely wrong. They just seemed easy and I was afraid I was making a simple mistake or something.

**Mathstud28** · November 14th 2008, 01:28 PM

Originally Posted by DemonGal711

I have to determine whether the sequence converges or diverges and, if it does converge, find the limit. I'm confused about these two.

arctan(2x)

The limit of arctan(x) as x goes to infinity is pi/2. So, doesn't arctan(2x) do the same, just the behavior around the origin is altered? Can I just put that the limit of arctan(2x) is pi/2 even if I really didn't do anything to prove it?

ln x
ln 2x

Both go to infinity, the denominator goes just a bit quicker. So, I do l'hopital rule and I get

1/x
2/2x

1/x * 2x/2

Which reduces to 1 if I'm not mistaken. So, I get the obvious limit.

lim 1 = 1

which means my original problem's limit is 1. Is that right? It just seemed to easy to me.

To really show these converges we must show that the sequences are bounded and monotonic. And then if they converge we can say the limit exists and find it.

1. Firstly let $a_n=\arctan(n)$ . Then $\sup\left(a_n\right)=\frac{\pi}{2}$ and $\inf\left(a_n\right)=\frac{-\pi}{2}$ . So then $\left|a_n\right|\leq\frac{\pi}{2}$ . Furthermore note that $a'(n)=\frac{1}{1+n^2}>0~\forall{n}\in\mathbb{N}$ . So we have show that $a_n$ is both bounded and monotonic. Thus it is convergent and its limit is $\lim_{n\to\infty}a_n=\frac{\pi}{2}$

2. Next let $a_n=\frac{\ln(n)}{\ln(2n)}=\frac{\ln(n)}{\ln(2)+\l n(n)}=\frac{1}{\frac{\ln(2)}{\ln(n)}+1}~\forall{n} \in\mathbb{N}>1$

So now consider that $\sup\left(a_n\right)=1$ and that $\inf\left(a_n\right)=\frac{1}{2}$ . So then $|a_n|\leq{1}$ . And notice that $\ln(n+1)>\ln(n)~\forall{n}\in\mathbb{N}>1\implies{ a_{n+1}>a_n}\forall{n}\in\mathbb{N}>1$ . Therefore we have shown that the sequence is bounded and monotonic, thus convergent. And then we state that $\lim_{n\to\infty}a_n=1$ .

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