1. ## Series...

Determine whether the series converges or diverges. If the series converges, find its limit. Justify your answers.

$\displaystyle \sum_{j=2}^{\infty} \frac{3^j}{4^{j+1}}$

So first I did this, although I'm not sure if I'm going in the right direction, although I believe I am...

= $\displaystyle \frac {3^2}{4^3} (1 + \frac {3}{4} + \frac {3^2}{4^2} + \frac {3^3}{4^3} ...)$ then I can't figure out what the stuff inside the parenthesis equals out to be, then I could figure out the answer.

2. Originally Posted by larson
Determine whether the series converges or diverges. If the series converges, find its limit. Justify your answers.

$\displaystyle \sum_{j=2}^{\infty} \frac{3^j}{4^{j+1}}$

So first I did this, although I'm not sure if I'm going in the right direction, although I believe I am...

= $\displaystyle \frac {3^2}{4^3} (1 + \frac {3}{4} + \frac {3^2}{4^2} + \frac {3^3}{4^3} ...)$ then I can't figure out what the stuff inside the parenthesis equals out to be, then I could figure out the answer.
Here I will give you big geometric hint...this can be rewritten as $\displaystyle \sum_{j=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}{ 4}\bigg)^{j}$

3. Originally Posted by Mathstud28
Here I will give you big geometric hint...this can be rewritten as $\displaystyle \sum_{j=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}{ 4}\bigg)^{j}$
is what I wrote correct though? Even if what you wrote is geometric, where does that put me?

4. Originally Posted by larson
is what I wrote correct though? Even if what you wrote is geometric, where does that put me?
its convergent since $\displaystyle \bigg|\frac{3}{4}\bigg|<1$ which is the definition of the convergent geometric series

5. Originally Posted by larson
is what I wrote correct though? Even if what you wrote is geometric, where does that put me?
$\displaystyle \sum_{n=0}^{\infty}r^n=\frac{1}{1-r}$ if r < 1

6. Originally Posted by Mathstud28
its convergent since $\displaystyle \bigg|\frac{3}{4}\bigg|<1$ which is the definition of the convergent geometric series
and the sum would be $\displaystyle \bigg(\frac{3}{4}\bigg)^2\cdot\bigg(\frac{1}{4}\bi gg)\cdot\frac{1}{1-\frac{3}{4}}$

7. Originally Posted by Mathstud28
and the sum would be $\displaystyle \bigg(\frac{3}{4}\bigg)^2\cdot\bigg(\frac{1}{4}\bi gg)\cdot\frac{1}{1-\frac{3}{4}}$
I also just figured it out the way I was first doing it by figuring out how to do series / sums on my TI-83 Plus . I figured out what was the in the parenthesis was 4, so $\displaystyle \frac {(4)3^2}{4^2} = \frac {9}{16}$.

8. Originally Posted by Mathstud28
Here I will give you big geometric hint...this can be rewritten as $\displaystyle \sum_{j=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}{ 4}\bigg)^{j}$
Could I ask you real fast how you knew to find that $\displaystyle \frac {1}{4}$

9. Originally Posted by larson
Could I ask you real fast how you knew to find that $\displaystyle \frac {1}{4}$
$\displaystyle 4^{j + 1} = 4 \cdot 4^j$. we wanted the $\displaystyle 4^j$, so we pulled the 4 out from the denominator making the $\displaystyle \frac 14$

10. Originally Posted by larson
Could I ask you real fast how you knew to find that $\displaystyle \frac {1}{4}$
sure you have $\displaystyle \sum_{n=2}^{\infty}\frac{3^n}{4^{n+1}}=\sum_{n=2}^ {\infty}\frac{3^n}{4\cdot{4^{n}}}=\sum_{n=2}^{\inf ty}\frac{1}{4}\cdot\frac{3^n}{4^n}$$\displaystyle =\sum_{n=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3} {4}\bigg)^n$

11. You may modify the sum by changing the lower limit to $\displaystyle 0.$ From there you'll get a direct geometric series.

12. Originally Posted by Krizalid
You may modify the sum by changing the lower limit to $\displaystyle 0.$ From there you'll get a direct geometric series.
You don't need to, you just have to say that :

$\displaystyle \sum_{k=m}^n x^k=x^m \frac{1-x^{m+n+1}}{1-x}$

Oh...I'm not sure I've understood correctly the thing actually

13. Originally Posted by Moo
You don't need to, you just have to say that :

$\displaystyle \sum_{k=m}^n x^k=x^m \frac{1-x^{m+n+1}}{1-x}$

Oh...I'm not sure I've understood correctly the thing actually
Actualy...I am pretty sure $\displaystyle \sum_{m=p}^{\infty}x^{n}=x^{p}\frac{1}{1-x},\forall{|x|}<1$

14. Of course...

I was just giving the general formula...

15. Originally Posted by Moo
Of course...

I was just giving the general formula...
OK!

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