The discussion revolves around the convergence of the infinite series represented by the sigma notation Ʃ 2^n/(4^n+1). Participants explore the application of the root test and the ratio test to determine convergence, while addressing potential misinterpretations of the series' structure.
The conversation includes various interpretations of the series and the tests applied. Some participants express confusion over the algebraic manipulations, while others provide clarifications. There is no explicit consensus on the final conclusion regarding convergence, but guidance on the ratio test has been offered.
Participants note the importance of parentheses in mathematical expressions and the potential for misunderstanding when they are omitted. There is also mention of the original poster's lack of familiarity with the ratio test due to missing class instruction.
Ʃ 2^n/(4^n+1)
rewrite as,
Ʃ 2^n/(4^n+1^n)
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}}##
SammyS said:OP was correct !
Actually, Ʃ 2^n/(4^n+1) is \displaystyle \ \sum\frac{2^n}{4^n+1}\ .
Yes. That's fine, but I don't see why that helps.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.?
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?
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.