1. ## series again!

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:
$\displaystyle \sum_{n=1}^{\infty}\frac{(n+1)}{n^3}x^n$

2. Investigate the convergence of the series

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

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!

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

3. ## re:

I wanted to know say you had:

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

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

How would you work out $\displaystyle U_n$.

Thank you

4. does it require a general formula?

5. Hello, Enita!

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

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

How would you work out $\displaystyle U_n$?

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

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

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

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

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

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

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

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

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

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

Hope this helps . . .