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Determine Whether series converge or diverge if it converge what is the limit.
$\{ \frac {(-1)^{n-2}n^{2}}{4+n^{3}}\} _{n=0}^{\infty }$
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- I have done a bunch of these of problems, but this one step looks wrong in the solution for this problem. https://www.bartleby.com/solution-answer/chapter-62-problem-71e-calculus-volume-2-17th-edition/9781938168062/in-the-following-exercises-express-each-series-as-a-rational-function-71-n11-x3-2n1/b209f360-2097-11e9-8385-02ee952b546e On the very first step when I distribute the (x-3)^(2n-1) exponent in the denominator to be (x-3)^2n and (x-3)^-1, the (x-3)^-1, should flip up to the numerator (not stay in the denominator as (x-3)), but it doesn't according to the answer key. And the answer is right but that first step LOOKS wrong. Can you help explain that exponent distribution?a_{n}=\frac{(-3)^{n}}{n !}Choose the true statements about a convergent series.
- Determine whether the sequence converges or diverges. If it converges, find the limit. (If the sequence diverges, enter DIVERGES.) an = n^3/ 3n + 9lim n→∞ an =Please explain why sequence Xnk approaches +infinity and why sequence Xmk approaches -infinity. Why would one use each of the sequences if the main sequence is (n sin n)? Lastly, can I assume sequence sequence Nk sin Nk <1/2 ...therefore making it convergent? Why the other way?Does the sequence {an} converge or diverge? Find the limit if the sequence is convergent. n = = (₁ + ²/)" Start by finding the natural log of an without using exponents. In an =