# ratio test.

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• Apr 8th 2007, 06:31 AM
ratio test.
Hi all!

I been working on series test, I wanted to ask about a step which I don't get. Heres the series:

1/5 + 2/6 + 2^2/7 + 2^3/8 + 2^4/9 + ...

applying the ratio test:

Un = 2^(n-1)/(4+n), Un+1 = 2^n/(5+n)

so therefore, Un+1 / Un = 2^n/(5+n).(4+n)/2^(n-1)

using the rule 2^(n-1) = 2^n.2^-1

the 2^n cancels leaving 2^-1, but I don't know how to get:

Un+1 / Un = 2(4+n)/(5+n)

------------------------------------

I'm having the same problem with another question to where:

Un = 2n-1/2^(n-1), Un+1 = 2n+1/2^n

Un+1 / Un = 2n+1/2^n.2^(n-1)/2n-1

I don't get how to get

Un+1 / Un = 1/2.2n+1/2n-1

Thank you.
• Apr 8th 2007, 07:08 AM
Soroban

Quote:

Here's the series: .1/5 + 2/6 + 2²/7 + 2³/8 + ...

Applying the ratio test:

. . U
n .= .2^(n-1)/(4+n), . Un+1 .= .2^n/(5+n)

so: .(U
n+1)/(Un) .= .2^n/(n + 5) · (n + 4)/2^(n-1) . Good!

but I don't know how to get: .(U
n+1)/(Un) .= .2(n + 4)/(n + 5)

I assume you mean you don't know how to find the limit.

Divide top and bottom by n, and we have:

. . . . 2(1 + 4/n) . . . .2(1 + 0)
. lim .------------ . = . ---------- . = . 2
n→∞ . 1 + 5/n . . . . . .1 + 0

• Apr 8th 2007, 07:14 AM
Hey!

Thanks for you reply. I wanted to know where the 2 comes from in front of the bracket (n+4).

and for the second question, where the 1/2 comes from.

Thank you
• Apr 8th 2007, 07:27 AM
CaptainBlack
Quote:

Hey!

Thanks for you reply. I wanted to know where the 2 comes from in front of the bracket (n+4).

and for the second question, where the 1/2 comes from.

Thank you

2=1/2^{-1}

and

1/2=2^{-1}

RonL
• Apr 8th 2007, 08:15 AM
Re:
Thanks for the reply. I think am just having problems simplifying:

Un+1 / Un = 2^n/(5+n).(4+n)/2^(n-1)

I don't get why you can cancel the 2^n but not the n.

and the same for:

Un+1 / Un = 2n+1/2^n.2^(n-1)/2n-1

I don't get why you can cancel the 2^n again and not the 2n.

I hope you can tell what the colours mean. lol

Thank you

(sorry I know this is a different question then the one asked at the start of the thread)

I have attached pic for the first example mentioned as it is more clear.
• Apr 8th 2007, 10:04 AM
earboth
Quote:

Thanks for the reply. I think am just having problems simplifying:

Un+1 / Un = 2^n/(5+n).(4+n)/2^(n-1)

I don't get why you can cancel the 2^n but not the n.

I've attached an image to show you how to simplify this term:
• Apr 8th 2007, 10:07 AM
Jhevon
Quote:

Thanks for the reply. I think am just having problems simplifying:

Un+1 / Un = 2^n/(5+n).(4+n)/2^(n-1)

I don't get why you can cancel the 2^n but not the n.

and the same for:

Un+1 / Un = 2n+1/2^n.2^(n-1)/2n-1

I don't get why you can cancel the 2^n again and not the 2n.

I hope you can tell what the colours mean. lol

Thank you

(sorry I know this is a different question then the one asked at the start of the thread)

I have attached pic for the first example mentioned as it is more clear.

a law of exponents says that if you have the same bases, one dividing the other, then the answer is given by the base raised to the power obtained by subtracting the power in the denomenator from the power in the numerator. that is,

x^m/x^n = x^(m-n)

similarly, we had a 2^n in the top and a 2^(n-1) in the bottom, so the result will be 2^(n - (n - 1)) = 2^1, which is what we are left with. since the power is positive we put it in the top

the n's can't cancel because they are part of a sum (i include subtractions here). you can only cancel when you have the top and bottom as a product (i include divisions here) where what you want to cancel can be factored out.
• Apr 8th 2007, 10:13 AM
earboth
Quote:

Thanks for the reply. I think am just having problems simplifying:

and the same for:

Un+1 / Un = 2n+1/2^n.2^(n-1)/2n-1

I don't get why you can cancel the 2^n again and not the 2n.

here is the image with the simplification of the second term:
• Apr 9th 2007, 02:52 AM

I will post an example for you to check.

Thanks.
• Apr 9th 2007, 12:18 PM
re:
I just wanted you to check one last question which I did to see if it is right.

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

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

(Un + 1) / (Un) = 1 + 2n^2 + 4n + 2 / 1 + n^2 + 2n + 1 . 1 + n^2 / 1 + 2n^2

(Un + 1) / (Un) = 2n^2 + 4n + 3 / n^2 + 2n + 2 . 1 + n^2 / 1 + 2n^2

cancelling out now to give:

(Un + 1) / (Un) = 4n + 3 / 2n + 2

Lim n → ∞
(Un + 1) / (Un) = 4 + (3/n) / 2 + (2/n)

= 4 + 0 / 2 + 0

= 2

as > 1 series diverges.

Thanks.
• Apr 9th 2007, 01:47 PM
Soroban

Quote:

Un .= .(1 + 2n²)/(1 + n²)

U
n+1 .= .[1 + 2(n+1)²]/[1 + (n+1)²]

(U
n+1 /(Un) .= .(1 + 2n² + 4n + 2)/(1 + n² + 2n + 1) · (1 + n²)/(1 + 2n²)

(U
n+1)/(Un) .= .(2n² + 4n + 3)/(n² + 2n + 2) · (1 + n²)/(1 + 2n²) . Good!

cancelling out now to give:
. . (U
n+1)/(Un) .= .(4n + 3)/(2n + 2)

But what did you cancel? . . . and how?

. . . . . . . . (2n² + 4n + 3)(n² + 1)
We have: . ---------------------------
. . . . . . . . (n² + 2n + 2)(2n² + 1)

Divide top and bottom by n^4

. . .2n² + 4n + 3 . . n² + 1
. . .---------------- . .---------
. . . . . . . . . . . . . . . . . . . . .(2 + 4/n + 3/n²)(1 + 1/n²)
. . ----------------- . ----------- · = . --------------------------------
. . .n² + 2n + 2 . . .2n² + 1 . . . . .(1 + 2n + 2/n²)(2 + 1/n²)
. . .--------------- . . ----------
. . . . . . . . . . . . . .

Now take the limit . . .

• Apr 9th 2007, 02:33 PM
Quote:

Originally Posted by Soroban
(Un+1)/(Un) = (4n + 3)/(2n + 2)

But what did you cancel? . . . and how?

I cancelled the 2n^2 and n^2 from both side. I assume I did it wrong :confused:
• Apr 10th 2007, 11:27 AM
is the limit 2?

thank you
• Apr 10th 2007, 11:27 AM
Jhevon
Quote:

is the limit 2?

thank you

yes

if the highest power in the top is equal to the highest power in the bottom, the limit is always the ratio of the coefficients of the highest power of the variable. you can divide each term by n^2 to see this
• Apr 10th 2007, 11:31 AM