Convergence/Divergence of Series.

panda* · November 9th 2008, 01:19 AM

I am really poor in this part, been trying to do as much exercises as possible to get familiar with the patterns and I realised there are these few patterns that I couldn't seem to approach it.

Sometimes I know the right approach for it, but I have troubles simplifying the terms in order to prove that it fits certain criteria required to do conclusions. Here are a few cases, I hope that you guys would be able to give me a guidance on how to approach these different cases.

Case I : When there is root of something involved.

(i) $\sum^\infty_{n=1} \frac{1}{\sqrt{2n+1}}$
(ii) $\sum^\infty_{n=1} \frac{1}{\sqrt[3]{n^2 + 2}}$
(iii) $\sum^\infty_{n=1} \frac{1}{\sqrt[3]{3n^4 - 2}}$

Case II : When there is to the power of n.

(i) $\sum^\infty_{n=0} \frac {1 + 3^n}{1 + 4^n}$
(ii) $\sum^\infty_{n=1} \frac {3^n + 7n}{2^n(n^2 +1)}$

I reckon you approach this one with the ratio test? I always have troubles simplifying them!

Case III : When there is polynomials.

(i) $\sum^\infty_{n=1} \frac{n^3 + 4n}{n^4 + 200}$
(ii) $\sum^\infty_{n=1} \frac{5n^3 + 14n}{2n^6 + 200n^2 - 7}$

And the following few questions, I don't know how to classify them.

(i) $\sum^\infty_{n=0} (-1)^n$ - Alternating Series??
(ii) $\sum^\infty_{n=1} log (\frac {(n+1)^2}{n(n+2)})$
(iii) $\sum^\infty_{n=1} \frac{(n!)^2}{(2n)!}$ - I suppose you use Ratio test? But I have troubles simplifying this one!

Quite a number of questions there, I know! So sorry for the overload of questions

Thanks for looking and helping though!!

Your help to master this topic would be greatly greatly appreciated!

Thank you so much!

**mr fantastic** · November 9th 2008, 01:51 AM

Originally Posted by panda*

I am really poor in this part, been trying to do as much exercises as possible to get familiar with the patterns and I realised there are these few patterns that I couldn't seem to approach it.

Sometimes I know the right approach for it, but I have troubles simplifying the terms in order to prove that it fits certain criteria required to do conclusions. Here are a few cases, I hope that you guys would be able to give me a guidance on how to approach these different cases.

Case I : When there is root of something involved.

(i) $\sum^\infty_{n=1} \frac{1}{\sqrt{2n+1}}$
(ii) $\sum^\infty_{n=1} \frac{1}{\sqrt[3]{n^2 + 2}}$
(iii) $\sum^\infty_{n=1} \frac{1}{\sqrt[3]{3n^4 - 2}}$

[snip]

Use the comparison test for all of them.

Note that:

(i) $\frac{1}{\sqrt{2n + 1}} > \frac{1}{\sqrt{2n + n}} = \frac{1}{\sqrt{3n}} = \frac{1}{\sqrt{3}} \, \frac{1}{n^{1/2}}$ .

(ii) $\frac{1}{\sqrt[3]{n^2 + 2}} > \frac{1}{\sqrt[3]{n^2 + 2n^2}} = \frac{1}{\sqrt[3]{3n^2}} = \frac{1}{\sqrt[3]{3}} \, \frac{1}{n^{2/3}}$ .

(iii) $\frac{1}{\sqrt[3]{3n^4 - 2}} < \frac{1}{\sqrt[3]{3n^4 - 2n^4}} = \frac{1}{\sqrt[3]{n^4}} = \frac{1}{n^{4/3}}$ .

These inequalities should make it obvious what series to compare with.

**mr fantastic** · November 9th 2008, 02:23 AM

Originally Posted by panda*

[snip]
Case III : When there is polynomials.

(i) $\sum^\infty_{n=1} \frac{n^3 + 4n}{n^4 + 200}$
(ii) $\sum^\infty_{n=1} \frac{5n^3 + 14n}{2n^6 + 200n^2 - 7}$

[snip]

Use the comparison test for each.

Note that:

(i) $\frac{n^3 + 4n}{n^4 + 200} > \frac{n^3}{n^4 + 200 n^4} = \frac{n^3}{201 n^4} = \frac{1}{201} \cdot \frac{n^3}{n^4} = \frac{1}{201} \cdot \frac{1}{n}$ .

(ii) $\frac{5n^3 + 14n}{2n^6 + 200n^2 - 7} < \frac{5n^3 + 14n^3}{2n^6 + 200n^2 - 7n^2} = \frac{19n^3}{2n^6 + 193n^2} < \frac{19n^3}{2n^6} = \frac{19}{2} \cdot \frac{n^3}{n^6} = \frac{19}{2} \cdot \frac{1}{n^3}$ .

**Moo** · November 9th 2008, 02:29 AM

To add to Mr F's great hints, here is something :

How can you know by just one sight that it converges or diverges ? How can you know if you have to find a diverging sequence that is inferior or a converging sequence that is superior ?

Study the asymptotic equivalence. This may sound a rude word, but here it is. Let's take $\frac{1}{\sqrt[3]{n^2+2}}$

When n goes very large, 2 can be neglected. More formally, you can write : $\lim_{n \to \infty} n^2+2=\lim_{n \to \infty} n^2$

So when n goes large, $\frac{1}{\sqrt[3]{n^2+2}} \sim \frac{1}{\sqrt[3]{n^2}}=\frac{1}{n^{2/3}}$

Now, you should know you have to compare it to the Riemann series :
$\sum_{n=1}^\infty \frac{1}{n^s}$ converges if and only if $s>1$

Is $\tfrac 23 > 1$ ? No. Hence it diverges.

When it comes to stuff with $n^4+3n$ , remember that when n is very large, the most powerful stays alive :
$n^4+3n \sim n^4$

**mr fantastic** · November 9th 2008, 02:37 AM

Originally Posted by panda*

[snip]
Case II : When there is to the power of n.

(i) $\sum^\infty_{n=0} \frac {1 + 3^n}{1 + 4^n}$
(ii) $\sum^\infty_{n=1} \frac {3^n + 7n}{2^n(n^2 +1)}$

[snip]

More comparison test.

Note that:

(i) $\frac {1 + 3^n}{1 + 4^n} < \frac {3^n + 3^n}{4^n} = 2 \, \left( \frac{3^n}{4^n} \right) = 2 \, \left( \frac{3}{4}\right)^n$ .

(ii) $\frac{3^n + 7n}{2^n(n^2 +1)} > \frac{3^n}{2^n (n^2 + n^2)} = \frac{3^n}{2^n (2 n^2)} = \frac{1}{2} \, \left( \frac{3^n}{n^2 \, 2^n} \right)$ .

And by the ratio test $\sum_{n=1}^{\infty} \frac{3^n}{n^2 \, 2^n}$ is known to be divergent.

**Mathstud28** · November 9th 2008, 07:08 AM

(i) $\sum^\infty_{n=0} (-1)^n$ - Alternating Series??

No need for Leibniz's Criterion since this series fails the most rudimentary test.

$\lim_{n\to\infty}(-1)^n\ne{0}$ therfore it diverges by the n-th term test.

(ii) $\sum^\infty_{n=1} log (\frac {(n+1)^2}{n(n+2)})$

If we write this as

$\sum_{n=1}^{\infty}\bigg[2\ln(x+1)-\ln(x)-\ln(x+2)\bigg]$

We realize that this is a telescoping series with sum $\ln(2)$

(iii) $\sum^\infty_{n=1} \frac{(n!)^2}{(2n)!}$ - I suppose you use Ratio test? But I have troubles simplifying this one!

I prefer root test, but ratio test is acceptable here.

$\begin{aligned}\lim_{n\to\infty}\bigg|\frac{(n+1)^ 2(n!)^2}{(2n+2)(2n+1)(2n)!}\cdot\frac{(2n)!}{(n!)^ 2}\bigg|&=\lim_{n\to\infty}\bigg|\frac{(n+1)^2}{(2 n+2)(2n+1)}\bigg|\\<br /> &=\frac{1}{4}<1<br /> \end{aligned}$

So the series converges

**Moo** · November 9th 2008, 07:29 AM

Originally Posted by Mathstud28

I prefer root test, but ratio test is acceptable here.

While dealing with factorials, in 99% of the cases, the ratio test is more straighforward.

**Mathstud28** · November 9th 2008, 07:54 AM

Originally Posted by Moo

While dealing with factorials, in 99% of the cases, the ratio test is more straighforward.

Which is exactly why I don't like it. =D

**Moo** · November 9th 2008, 08:13 AM

Originally Posted by Mathstud28

Which is exactly why I don't like it. =D

Hmmm, so why do you like L'Hôpital's rule ?

**Mathstud28** · November 9th 2008, 08:32 AM

Originally Posted by Moo

Hmmm, so why do you like L'Hôpital's rule ?

I don't anymore. You converted me. I wrote a 175 page treatise on limit techniques, none of which are L'hopital's.

**Moo** · November 9th 2008, 08:35 AM

Originally Posted by Mathstud28

I don't anymore. You converted me. I wrote a 175 page treatise on limit techniques, none of which are L'hopital's.

Maybe I can write a 175 letters text on why it's pointless to use the root test when the ratio test works perfectly, avoiding to use any Stirling approximations or whatever unknown limits.
But it would not be a nice text

That's good you don't like l'Hospital's rule anymore

panda* · November 14th 2008, 04:58 PM

Thank you for all the help you guys

I have took your guidance and attempted the questions and I am glad to say that I am finally getting the hang of it!

Hopefully I'd be able to ace this topic in the exams!

Thank you once again!

**mr fantastic** · November 14th 2008, 06:33 PM

Originally Posted by panda*

Thank you for all the help you guys

I have took your guidance and attempted the questions and I am glad to say that I am finally getting the hang of it!

Hopefully I'd be able to ace this topic in the exams!

Thank you once again!

It's a pleasure to help someone who's so appreciative and with such lovely manners.

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