Question about root test for series

In summary, the given sigma notation for an infinite series, Ʃ 2^n/(4^n+1), can be rewritten as Ʃ 2^n/(4^n+1^n). However, this is incorrect and the correct series is actually Ʃ 2^n/(4^(n+1)). The root test can be used by taking the limit as n approaches infinity of n^(√(abs((2/(4+1))^n))). This limit evaluates to 2/5, and since it is less than 1, the series converges.
  • #1
banshee43
15
0

Homework Statement



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.
 
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  • #2
Ʃ 2^n/(4^n+1)
rewrite as,

Ʃ 2^n/(4^n+1^n)

No this is 100% wrong to do.

Your series is :

##\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
sorry i didnt put enough parentheses my series is 2^n/((4^n)+1)
 
  • #4
Zondrina said:
banshee43 said:

Ʃ 2^n/(4^n+1)
rewrite as,

Ʃ 2^n/(4^n+1^n)



No this is 100% wrong to do.

Your series is :

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

Which is definitely NOT equal to :

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

OP was correct !

Actually, Ʃ 2^n/(4^n+1) is [itex]\displaystyle \ \sum\frac{2^n}{4^n+1}\ .[/itex]
 
  • #5
SammyS said:
OP was correct !

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

Sorry, the lack of parenthesis confused me.
 
  • #6
So can that 1 be represented by a 1^n? Because no matter what n is Given n>=0 that 1^n = 1.?
 
  • #7
banshee43 said:
So can that 1 be represented by a 1^n? Because no matter what n is Given n>=0 that 1^n = 1.?
Yes. That's fine, but I don't see why that helps.

You do realize that [itex]\displaystyle \ 4^n+1^n\ne(4+1)^n\,,\ [/itex] don't you ?
 
  • #8
I thought that it looked funky when I did it. Thank you though
 
  • #9
So, do you have a solution ?
 
  • #10
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
banshee43 said:
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 algebra
(4^(n)+1)/(4*4^(n)+1) and some more algebra...
gives me 1/4 which is <1 therefore the series converges?

Are you using a limit without stating that you are?

[itex]\displaystyle \frac{4^n+1}{4^{n+1}+1}\ne \frac{1}{4}\ .[/itex]

Also, what happened to the 2 ?
 
  • #12
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
banshee43 said:
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 wasn't in class when we learned it.

Anyway, the series converges.
 

1. What is the root test for series?

The root test for series is a convergence test that is used to determine whether a series converges or diverges by comparing it to a geometric series. It is also known as the Cauchy's root test or the nth root test.

2. How does the root test for series work?

The root test for series works by taking the nth root of the absolute value of the terms in a series and then comparing it to a geometric series with a common ratio less than 1. If the limit of the nth root is less than 1, the series converges. If it is greater than 1, the series diverges. And if it is equal to 1, the test is inconclusive.

3. When should the root test for series be used?

The root test for series should be used when the series contains non-negative terms and the ratio test is inconclusive. It is also useful when the terms in the series involve factorials or other complicated expressions that make it difficult to apply other convergence tests.

4. Can the root test for series be used for all series?

No, the root test for series can only be used for series that contain non-negative terms. If the series contains negative terms, the root test cannot be applied. In such cases, other convergence tests such as the ratio test or the integral test should be used.

5. What is the difference between the root test and the ratio test?

The root test and the ratio test are both convergence tests, but they use different approaches to determine convergence. The root test involves taking the nth root of the terms in the series, while the ratio test involves taking the ratio of consecutive terms. The root test is generally used for series with factorials or other complicated expressions, while the ratio test is more versatile and can be used for a wider range of series.

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