what series test do you use for the sum starts at n=1 (n-1)/(n4^n)?

- anonymous

what series test do you use for the sum starts at n=1 (n-1)/(n4^n)?

- jamiebookeater

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- anonymous

\[\sum_{n=1}^{\infty}\]\[(n-1)\(n4^n)\]

- anonymous

n=(1(n-1))/(n^(4^(n)))
Since n is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
(1(n-1))/(n^(4^(n)))=n
Reduce the expression (1(n-1))/(n^(4^(n))) by removing a factor of from the numerator and denominator.
(n-1)/(n^(4)n^(4))=n
Reduce the expression ((n-1))/(n^(4)n^(4)) by removing a factor of from the numerator and denominator.
(n-1)/(n^(8))=n
Find the LCD (least common denominator) of ((n-1))/(n^(8))+n.
Least common denominator: n^(8)
Multiply each term in the equation by n^(8) in order to remove all the denominators from the equation.
(n-1)/(n^(8))*n^(8)=n*n^(8)
Simplify the left-hand side of the equation by canceling the common terms.
n-1=n*n^(8)
Multiply n by n^(8) to get n^(9).
n-1=n^(9)
Since n^(9) contains the variable to solve for, move it to the left-hand side of the equation by subtracting n^(9) from both sides.
n-1-n^(9)=0
Move all terms not containing n to the right-hand side of the equation.
-n^(9)+n-1=0
Divide each term in the equation by -1.
n^(9)-n+1=0
The Durand-Kerner method does not converge for this polynomial.
No Convergence

- anonymous

that what you wanted

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## More answers

- anonymous

Well we haven't learned that method yet the only ones we can use are geometric series, telescoping series, p series, divergence test, direct comparison test, and limit comparison test......

- anonymous

also i know from the answer in the back it is convergent...

- anonymous

so isnt what want to know right ?

- anonymous

i wanna know if it converges or diverges, then we must show using one of the tests above to prove it..

- anonymous

solving for n or x

- anonymous

n

- anonymous

No Convergence
use the Durana-Kerner

- anonymous

way i remember it

- anonymous

The factorial function is formally defined by
or recursively defined by
Both of the above definitions incorporate the instance
in the first case by the convention that the product of no numbers at all is 1. This is useful because:
There is exactly one permutation of zero objects (with nothing to permute, "everything" is left in place).
The recurrence relation (n + 1)! = n! × (n + 1), valid for n > 0, extends to n = 0.
It allows for the expression of many formulas, like the exponential function as a power series:
It makes many identities in combinatorics valid for all applicable sizes. The number of ways to choose 0 elements from the empty set is . More generally, the number of ways to choose (all) n elements among a set of n is .
The factorial function can also be defined for non-integer values using more advanced mathematics, detailed in the section below. This more generalized definition is used by advanced calculators and mathematical software such as Maple or Mathematica.

- anonymous

In mathematics, the factorial of a positive integer n,[1] denoted by n!, is the product of all positive integers less than or equal to n. For example,

- anonymous

5!=5X4X3X2X1=120

- anonymous

The first series is ∑1/4^n, which is convergent: 1/4 + 1/16 + 1/64 + ... (It converges to 1/3, by the way.) The second series is ∑1/n*4^n, which also has to be convergent, since each term after the first one is smaller than the corresponding term in the first series. So, you have the sum (or difference, if you prefer) of two convergent series, which is therefore convergent

- anonymous

that help you anymore

- anonymous

more i mean hopefuly it did i tried to help

- anonymous

ya it did i figured its like using the limit comparison test..

- anonymous

thanks =)

- anonymous

okay just asking so information i gave you helped you solve it

- anonymous

no problem any time i can help

- anonymous

Do you know what test to use to determine if this is convergent or divergent...\[\sum_{n=1}^{\infty} \sin (1\n)\]

- anonymous

We know that 0 < (n - 1) < n, ∀n ≥ 2.
Hence √(n)/(n - 1) > √(n)/n > 0, ∀n ≥ 2.
But √(n)/n = 1/√(n) = 1/[n^(1/2)],
so √(n)/(n - 1) > 1/[n^(1/2)] > 0, ∀n ≥ 2.
Σ{n=2 to ∞}{1/[n^(1/2)]} is a p-series with p = 1/2 ≤ 1, so it diverges.
Thus Σ{n=2 to ∞}{√(n)/(n - 1)} also diverges by the comparison test

- anonymous

what does that upside down A mean?

- anonymous

The Divergence Test: If limn!1 an 6= 0 or does not exist then
P1
n=1 an is divergent.
The Integral Test: If f(x) is continous, positive and decreasing on [1;1] and f(n) = an
then
P1
n=1 an is convergent if and only if
R 1
1 f(x)dx is convergent.
The Comparison Test: If an 0 and bn 0 for all n then:
(a) if
P
bn is convergent and an bn for all n, then
P
an is also convergent
(b) if
P
bn is divergent and an bn for all n, then
P
an is also divergent
The Limit Comparison Test: If an 0 and bn 0 for all n and limn!1
an
bn
= c > 0
then either both
P1
n=1 an and
P1
n=1 bn are convergent or both are divergent.
The Alternating Series Test: Suppose we have
P1
n=1(1)nbn and bn > 0. If bn+1 bn
for all n and limn!1 bn = 0 then the series
P1
n=1(1)nbn is convergent.
The Ratio test: If
lim
n!1
an+1
an
=
8>< >:
L > 1 the series
P1
n=1 an is divergent;
L < 1
P1
n=1 an is absolutely convergent;
= 1 the Ratio Test is inconclusive
Two things good to know:
Def. We say
P1
n=1 an is absolutely convergent if
P1
n=1 janj is convergent.
Thm. If
P1
n=1 an is absolutely convergent then it is convergent.
1

- anonymous

umm what is all that about? second where did you pull that square root out of?

- anonymous

We would like to show that
(1)
1X
n=1
(1)n 2n2 + 3
(n + 1)(n + 3)
is divergent
(2)
1X
n=1
ln(n)
n + 1
is divergent
(3)
1X
n=1
en is convergent
(4)
1X
n=1
sin(n)
n2 is absolutely convergent. Is it convergent?
(5)
1X
n=1
(1)n n + 1
n2 is convergent but not absolutely convergent
(6)
1X
n=1
(1)n n3n!
3n is divergent
us

- anonymous

that help any

- anonymous

ummmmm give me a fews...

- anonymous

not really but i can ask the prof tomorrow... would you like to try another?? \[\sum_{1}^{\infty} n(e ^{-n)}\]

- anonymous

well i am trying lol

- anonymous

ya i know thanks =)

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