1. ## seriess

hello everyonee!

i need help on these questions please.

1) determine wether each of the followin series is convergent:

a) 2/2^2 + 2^2/2^2 + 2^3/3^2 + ... 2^n/n^2

b) Un = 1+2n^2/1+n^2

2)

does this series in picture converge (sorry did not know how to write in text)

2. Originally Posted by Enita
a) 2/2^2 + 2^2/2^2 + 2^3/3^2 + ... 2^n/n^2
The general form is,
$\sum_{n=1}^{\infty} \frac{2^n}{n^2}$
Use the ratio test,
$\left| \frac{2^{n+1}}{(n+1)^2} \cdot \frac{n^2}{2^n} \right| =\left| \frac{2n^2}{n^2+2n+1}\right|=\left| \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}\right|$
The limit is 2 as $n\to \infty$.
Thus, the ratio test says the series diverges.

3. can you explain the ratio test method please?

what the last line mean? starting with C

4. Originally Posted by Enita

b) Un = 1+2n^2/1+n^2
The limits as $n\to \infty$ is not equal to zero.
Thus it diverges.

5. Originally Posted by Enita
can you explain the ratio test method please?
I took,
$a_{n+1}=\frac{2^{n+1}}{(n+1)^2}$
And
$a_n=\frac{2^n}{n^2}$
Then divided them, (flip fractions)
And also used the fact that,
$2^{n+1}=2\cdot 2^n$.

what the last line mean? starting with C
That is the factor ring formed.... ignore that. It is my signature.

6. oh ok

i don't get this bit:
$=\left| \frac{2n^2}{n^2+2n+1}\right|=\left| \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}\right|$
The limit is 2 as $n\to \infty$.

what if the limit was different?

7. Originally Posted by Enita
oh ok

i don't get this bit:
$=\left| \frac{2n^2}{n^2+2n+1}\right|=\left| \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}\right|$
The limit is 2 as $n\to \infty$.

what if the limit was different?
Divide the numerator and denominator by $n^2$.

8. hey thanks alot!

I been working on these today and came across a example on the internet. But there is a process that i don't understand. I have attached a pic from the site and labelled what i don't understand with a red arrow. Hopefully if someone could help i would be very greatful.

how does it cancel?

(site: Calculus II (Math 2414) - Series & Sequences - Ratio Test)

thanxs

9. Okay,
He expresses,
$4^{2n+1}=4^{2n}\cdot 4$ in the numerator.
And,
$4^{2n+3}=4^{2n}\cdot 4^3$ in the denominator.
After some canceling he gets,
$\frac{4}{4^3}=\frac{1}{4^2}$

10. Hello, Enita!

Let's work with the fraction
. . and ignore the limit and absolute value for now.

We have: . $\frac{(-10)^{n+1}} {4^{2n+3}(n+2)} \cdot\frac{4^{2n+1}(n + 1)}{(-10)^n} \;=\;\frac{(-10)^{n+1}}{(-10)^n} \cdot \frac{4^{2n+1}}{4^{2n+3}} \cdot\frac{n+1}{n+2}$

Simplify: . $\frac{-10}{1}\cdot\frac{1}{4^2}\cdot\frac{n+1}{n+2} \;=\;\frac{-10(n+1)}{16(n+2)}$ . . . see?

11. thanks. you guys are great! just like having a teacher right here.

Thank you!

12. Originally Posted by Enita
thanks. you guys are great! just like having a teacher right here.

13. does anyone know how to solve question 2?

thanks math forum

14. Originally Posted by Enita
does anyone know how to solve question 2?

thanks math forum
That one is simple it is bounded by,
$a_n=\frac{1}{n!}$
And $\sum a_n$ is known to converge.

15. ## re:

I worked this out for $Un = \frac{1 + 2n^2}{1+n^2}$

$\left| \frac{1+2(n+1)^2}{1+(n+1)^2} \cdot \frac{1+n^2}{1+2n^2} \right| = \left| \frac{1+2(n^2+2n+1)}{1+(n^2+2n+1)} \cdot \frac{1+n^2}{1+2n^2} \right|$

$=\left| \frac{1+2n^2+4n+2}{1+n^2+2n+1} \cdot \frac{1+n^2}{1+2n^2} \right|=\left| \frac{4n+2}{2n+1}\right|$

divde by n
$=\left| \frac{4+2}{2+1}\right|
$

Series equals 6/3 which is > 1 so diverges?

Have i worked this out right? (using the ratio test and canceling?)

Thank you maths forum.

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