sumation notation n=1 to infinity of x^n/(n)^1/5    (5th root). Find the radius of convergence?
365360
Use the ratio test. We want the absolute value of the ratio as k approaches infinity of the (k+1)th term to the kth term to be less than 1. So take the expression x^n/n^(1/5) and let n = k + 1. Stick the result in the numerator. Then let n = k and stick that result in the denominator. Now take the limit as k goes to infinity. You should just end up with x (I don't want to write the whole process out here cuz it's hard to write equations but you should get something that looks like x[k/(k+1)]^(1/5) which approaches x as k -> infinity). So if |x| < 1, you're good to go. If x is plus or minus 1, we're still not sure if the series converges or not. The x = 1 case gives us a p-series with p = 1/5, which diverges. The x = -1 case gives us an alternating series in which the magnitude of each term is always decreasing and approaches zero as n goes to infinity. Thus, it passes the alternating series test for convergence and the series converges for x = -1.\n\nSo the interval of convergence is [-1,1) and the radius of convergence is 1.
388431
This looks like e^x = Sigma_{n=0}^{infty}x^n/n!, does it not?\n\nNow we want to make the signs alternate, so we try \ne^(-x) = Sigma_{n=0}^{infty}(-x)^n/n! \n\n= Sigma_{n=0}^{infty} (-1)^n*x^n/n!\n\nLastly, we want to get a 5 in the exponent but not in the denominator so we try \n\ne^(-x^5) = Sigma_{n=0}^{infty}(-x^5)^n/n! = \n\nSigma_{n=0}^{infty} (-1)^n*x^(5n)/n!\n\nTherefore the ****ANSWER is e^(-x^5).****  \n\n(NOT to be confused with e^(-5x))
890367
You divide the bottom into the top and then see what happens as the limit of x--> infinity.\n\nso x^3/(x^2-4) = x+4x/(x^2-4). Now notice that when x goes to infinity the second part of our answer is going to 0 because the x^2 part is going to dominate the 4x in the numerator. So when this function gets very large it is going to be closer and closer to y= x +0.\n\nThis is all that is meant by asymptotes. They are the functions that aproximate another function as x reacher infinity or a certain constant where the equation is undefined (0 in the denominator)\n\nhope this helps.
1023164
Limit ( n to infinity) (n + i/n)^ n
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
804449
There is no limit. The problem is that sin(x) goes between postive one and negative one infinitely many times as x gets small (i.e. goes to negative infinity). The hard part, though, is to figure out how to show this.\n\nI will look at two different sequences of numbers that go to negative infinity. When I plug the numbers in the first sequence into x/(2sin(x)) they will go to positive infinity. When I plug the numbers in the second sequence into x/(ssin(x)) they will go to negative infinity. If I can get two different answers then there is no limit.\n\nThe first sequence is x_n = -pi/2, -pi/2-2pi, -pi/2-4pi, -pi/2-6pi, ....,-pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-pi/2-2n pi)/(2sin(-pi/2-2n pi))=(-pi/2-2n pi)/(-2)=pi/2+2n pi\nAs n gets big then this number goes to positive infinity.\n\nThe second sequence is x_n = -3pi/2, -3pi/2-2pi, -3pi/2-4pi, -3pi/2-6pi, ....,-3pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-3pi/2-2n pi)/(2*sin(-3pi/2-2n pi))=(-3pi/2-2n pi)/(2)=-3pi/2-2n pi\nAs n gets big then this number goes to negative infinity.\n\nWe now have two sequences of numbers that give different answers. The function cannot converge to a single number.
871092
I guess you probably use the Epsilon definition for convergence? Then, you know:\nFor every Epsilon greater than 0, there exists an N such that n>N implies /xn - a/ < E'.\nYou want to show that this also implies:\n/xn^(1/3) - a^(1/3)/ < E.\n\nI think that you can probably do something with:\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\nmaking x = a in this equation, the proof's a = b.\nThen, you can find an N so that /[xn^(1/3) - a^(1/3)][xn^2/3 + xn^(1/3)a^(1/3) + a^(2/3)]/ < E'. Perhaps this is easier to relate to the first one?\n\nGood luck!
1136716
2r<22\nr<11\nAnswer= (-infinity,11)\n\n* don't have the infinity symbol on computer
767320
For the first one: yes, you can multiply the expression term by term. If f(x) is a function given by [n=0, ∞]∑(a_n x^n), and g(x) is any function, then f(x) * g(x) = [n=0, ∞]∑(g(x) a_n x^n). g(x) is just a constant in terms of n, so distributing it across the terms of the series will not change its convergence properties.\n\nFor the second question: you seem to be describing a Laurent series. These series are described in more detail on the Wikipedia:\n\nhttp://en.wikipedia.org/wiki/Laurent_series
284432
There is no sign for exponent in your question.  I assume you mean\n2^(3/5)\nAny expression of the from x^(m/n) can be written as the\nnth root of x^m\nThus, 2^(3/5) is the fifth root of 2^3 which simplifies to \nthe fifth root of 8.
510352
The solution by a_ebnlhaitham is not quite correct in that he omitted the 2 from the power of r (i.e.(teta)).The answer should be 1/(1 - r^2) not 1/(1 - r). The series converges because r lies in the open interval (-1,1).
766268
This series converges to \n(sqrt(3)*pi/4)-(4ln(2))+\n((9/4)ln(3)).
451467
The series converges because as n increases the ratio gets closer to .8. So A2=9/13 and A3=13/18 and so on. So each iteration makes the +1 in the numerator and the +3 in the denominator more and more negligible.So for n=∞, An=.8 assuming that the series is infinite.
49217
This is totally unfamiliar and I don't find it in my references.\n\nYou can prove that the series converges. \n\nn!/n^n = (1/n)(2/n)( n-2 other terms all < 1) < (1/n)(2/n)\nso an < 2/n^2. Since the sum(1/n^2) converges, your series converges. But that is as far as I am prepared to go.\n\nNote: If this were a Taylor series, since the nth term of the series is the nth derivative of f over n!, the nth derivative would have to have (n!)^2 in the numerator to cancel out the n! in the denominator.\n\nAs far as the answeres above, \na3 = (3*2*1) / (3^3) = 2/9, not 1/3.
19421
X^5 = 98\n=> ln(X^5) = ln98\n=> 5lnX = ln98\n=> lnX = (ln98)/5\n=> X = e^((ln98)/5)\nX = 2.50175752714084352...
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
81246
Let A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... \n\n(Note that A is equal to ln 2)\n\nA = (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 + ...) \n\nA = (1 + 1/3 + 1/5 + ...) - 1/2 (1 + 1/2 + 1/3 + ...) \n\nA = (1 + 1/3 + 1/5 + ...) - 1/2 (1 + 1/3 + 1/5 + ...) - 1/2 (1/2 + 1/4 +1/6 + ...) \n\nA = 1/2 (1 + 1/3 + 1/5 + ...) - 1/2 (1/2 + 1/4 +1/6 + ...) \n\nA = 1/2 (1 - 1/2 + 1/3 - 1/4 + 1/5 - ...) \n\nA = 1/2 A \n\n1 = 1/2 \n\n2 = 1 \n\n\n----------------------------------------------------------\nSpoiler - do not read below until you try to find the error\n-----------------------------------------------------------\n\nThe problem here is that the sequence, although it converges is said to be "conditionally convergent". The sequence as stated in the first line is considered to converge, by when split apart into even and odd members, has been transformed into an infinity minus infinity, thus it is no longer considered to converge.\n\nI used to play with these types of proofs in high school, and this was by far my favourite!