# Question about root test for series

1. Apr 6, 2013

### banshee43

1. The problem statement, all variables and given/known data

Can you if given a sigma notation for an infinite series

Ʃ 2^n/(4^n+1)
rewrite as,

Ʃ 2^n/(4^n+1^n)
then doing the root test can you

lim n→∞ n^√abs((2/(4+1))^n)

which equals 2/5, and
2/5<1, therefore can i conclude that the series above converges?

Sorry if this is in the wrong section, i guess its more of an algebra question.

2. Apr 6, 2013

### Zondrina

No this is 100% wrong to do.

$\sum \frac{2^n}{4^{n+1}}$

Which is definitely NOT equal to :

$\sum \frac{2^n}{4^{n} + 1^{n}}$

Why don't you try the ratio test, it will work perfectly for this.

3. Apr 6, 2013

### banshee43

sorry i didnt put enough parentheses my series is 2^n/((4^n)+1)

4. Apr 6, 2013

### SammyS

Staff Emeritus

OP was correct !

Actually, Ʃ 2^n/(4^n+1) is $\displaystyle \ \sum\frac{2^n}{4^n+1}\ .$

5. Apr 6, 2013

### Zondrina

Sorry, the lack of parenthesis confused me.

6. Apr 7, 2013

### banshee43

So can that 1 be represented by a 1^n? Because no matter what n is Given n>=0 that 1^n = 1.?

7. Apr 7, 2013

### SammyS

Staff Emeritus
Yes. That's fine, but I don't see why that helps.

You do realize that $\displaystyle \ 4^n+1^n\ne(4+1)^n\,,\$ don't you ?

8. Apr 7, 2013

### banshee43

I thought that it looked funky when I did it. Thank you though

9. Apr 7, 2013

### SammyS

Staff Emeritus
So, do you have a solution ?

10. Apr 7, 2013

### banshee43

Using the ratio test because there are constants raised to powers i get..
(2^(n+1)/(4^(n+1)+1))*((4^(n)+1)/2^n) then using some algerbra
(4^(n)+1)/(4*4^(n)+1) and some more algerbra....
gives me 1/4 which is <1 therefore the series converges?

11. Apr 7, 2013

### SammyS

Staff Emeritus
Are you using a limit without stating that you are?

$\displaystyle \frac{4^n+1}{4^{n+1}+1}\ne \frac{1}{4}\ .$

Also, what happened to the 2 ?

12. Apr 7, 2013

### banshee43

yes i am using a limit and i forgot about the 2, i guess it converges to 1/2 then? i had forgotten that the ratio test needed a limit because i wasnt in class when we learned it.

13. Apr 7, 2013

### SammyS

Staff Emeritus
Anyway, the series converges.