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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty}\]\[(n-1)\(n4^n)\]

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    that what you wanted

  4. anonymous
    • 5 years ago
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    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......

  5. anonymous
    • 5 years ago
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    also i know from the answer in the back it is convergent...

  6. anonymous
    • 5 years ago
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    so isnt what want to know right ?

  7. anonymous
    • 5 years ago
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    i wanna know if it converges or diverges, then we must show using one of the tests above to prove it..

  8. anonymous
    • 5 years ago
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    solving for n or x

  9. anonymous
    • 5 years ago
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    n

  10. anonymous
    • 5 years ago
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    No Convergence use the Durana-Kerner

  11. anonymous
    • 5 years ago
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    way i remember it

  12. anonymous
    • 5 years ago
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    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.

  13. anonymous
    • 5 years ago
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    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,

  14. anonymous
    • 5 years ago
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    5!=5X4X3X2X1=120

  15. anonymous
    • 5 years ago
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    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

  16. anonymous
    • 5 years ago
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    that help you anymore

  17. anonymous
    • 5 years ago
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    more i mean hopefuly it did i tried to help

  18. anonymous
    • 5 years ago
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    ya it did i figured its like using the limit comparison test..

  19. anonymous
    • 5 years ago
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    thanks =)

  20. anonymous
    • 5 years ago
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    okay just asking so information i gave you helped you solve it

  21. anonymous
    • 5 years ago
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    no problem any time i can help

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

  23. anonymous
    • 5 years ago
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    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

  24. anonymous
    • 5 years ago
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    what does that upside down A mean?

  25. anonymous
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    umm what is all that about? second where did you pull that square root out of?

  27. anonymous
    • 5 years ago
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    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

  28. anonymous
    • 5 years ago
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    that help any

  29. anonymous
    • 5 years ago
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    ummmmm give me a fews...

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

  31. anonymous
    • 5 years ago
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    well i am trying lol

  32. anonymous
    • 5 years ago
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    ya i know thanks =)

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