series again!

**Enita** · January 31st 2007, 01:19 PM

Hey! I need some help again with series. I hate this topic

lol

how would you do these problems

1. Finde the range of values for which the following series is convergent or divergent:
$\sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n$

2. Investigate the convergence of the series

$\frac{1}{1.2}+\frac{x}{2.3}+\frac{x^2}{3.4}+\frac{ x^3}{4.5}+...$ for $x>0$

sorry to flood you with questions, its because I have a test very soon on this.

and want to get a good mark

thank you!

**ThePerfectHacker** · January 31st 2007, 05:27 PM

Originally Posted by Enita

Hey! I need some help again with series. I hate this topic

lol

how would you do these problems

1. Finde the range of values for which the following series is convergent or divergent:
$\sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n$

Ratio test,
$\frac{n+2}{(n+1)^3}\cdot \frac{n^3}{n+1}$
I will leave you to check that the limit is 1.
Thus,
$|x|<1$
We have absolute convergence.
$|x|>2$
We have divergence.

**Enita** · February 3rd 2007, 07:58 AM

Thanks for your reply.

I wanted to know say you had:

1. $\frac{1}{1}+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3 }+...$

How do you know that:
$U_n=\frac{2n-1}{2^{n-1}}$

How would you work out $U_n$ .

Thank you

**Enita** · February 5th 2007, 10:03 AM

does it require a general formula?

**Soroban** · February 5th 2007, 02:02 PM

Hello, Enita!

I wanted to know, say you had: . $\frac{1}{1}+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3 }+ \cdots$

How do you know that: . $U_n\:=\:\frac{2n-1}{2^{n-1}}$

How would you work out $U_n$ ?

You do it by Inspection . . . and a bit of Experience.
. . You're expected to "eyeball" it.

The numerators are: $1,\,3,\,5,\,7,\,\cdots$
. . You recognize them as consecutive odd numbers.
And you're supposed to know the general formula: . $a_n \:=\:2n-1$

The denominators are: $1,\,2,\,2^2,\,2^3,\,\cdots$
. . which you should recognize as consecutive powers of 2.
Since it begins with $1$ , the first denominator must be $2^o$ .
. . So the denominators are: . $2^{n-1}$

Therefore: . $U_n \:=\:\frac{2n-1}{2^{n-1}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Another example: . $\frac{5}{9} + \frac{8}{27} + \frac{11}{81} + \frac{14}{243} + \frac{17}{729} + \cdots$

The numerators are: $5,\,8,\,11,\,14,\,17,\cdots$
They "go up by 3" . . . This tells us that the formula is: $3n + a$

The first term is $8.$
So when $n = 1$ , we have: . $3(1) + a \:=\:8\quad\Rightarrow\quad a \,=\,5$
. . Hence, the formula for the numerator is: . $\boxed{3n + 5}$

The denominators are: $9,\,27,\,81,\,243,\,729,\cdots$
. . They go up by a multiple of 3. .The formula is $3^k$
Since it begins with $9$ , the first denominator is $3^2.$
. . So the denominators are: . $\boxed{3^{n+1}}$

Therefore, the general term is: . $U_n \:=\:\frac{3n+5}{3^{n+1}}$

Hope this helps . . .

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