what is the ratio test for a series?
342283
The ratio test is a test for a series, basically to determine wether or not it converges. Roughly speak it looks at the ratio of two successive terms and determines wether this tends to zero (i.e. the series converges)\n\nHeres a complete guide to using it
664851
Dear, Your answers with complete explanation are below:\n\n1. Take any 2 terms, say 1/3 and 1/9.\nthe ratio is  = 1/3 / 1/9 = 1/3.\n\nSum of a geometric series to 'n' terms is given by:\nS(n) = a(r^n - 1)/(r - 1),\nwhere, \nS(n) = sum upto n terms\na= first term\nr=ration between 2 consecutive terms\nn= no. of terms\n\n2. Here,\n\nr= 1/3\nn= 10\na = 1.\n\nSo, S(10) = (3^10 - 1)x3 / 3^10 x 2.\nS(10) = 1.49997459 ~ 1.4999746\n\n3. Similarly, S(12) = 1.499997177\n\n4. If there are infinite terms in a geometric series, then its sum is = a/(1-r)\n\nHere this is = 1/(1-1/3) = 3/2 = 1.5\nSo, this series is always less than 1.5.
543595
Stirling's approximation:\n\nn! is approximately\n\nsqrt(2*pi*n)*(n/e)^n,\n\nso ln(n!) is about\n(n+1/2)*ln(n)-n +(1/2)ln(2pi)
823408
Your ratio is 1:8.15          You fid this out by starting at 13:106 and work yoiur way down to 1. \n\nyou divide both sides by 13 to get your answer.
948494
The correct term is "order statistic" which is the set of values arranged in order of magnitude.  Also a set of successive components with identical values though is called a "run".\n\nProgression is term used in sequences and series.
764231
Assuming that the first term in the geometric series is A, and that the ratio between any two successive terms is r, the first 12 terms of the series would be A+Ar+Ar^2+Ar^3+...+Ar^11.\n\nThe sum of the series is A[(r^12-1)/(r-1)]\n\nGood Luck :-)
831566
log(n)/n > 1/n if n>2, and the sum of 1/n diverges (this is the so-called harmonic series), so log(n)/n also diverges.
12336
Try the BBC's Maths series and MathsFile as well as Math.com (see links below)
289664
This ratio is called the cosine.
133118
There are formulas on the following page:\n\nhttp://www.acts.tinet.ie/trig.ratiosforacuteang_668.html
631314
so you've basically asked every possible series formula question here...but i'll still help you!  the formula for the sum of an arithmetic sequence is:\n\nS(n)=(n/2)[a(1)+a(n)]\n\nthe sume of first 20 terms is then:\n\nS(20)=(20/2)[a(1)+a(20)]\n\nso you need to know the first term, a(1), and the 20th term, or a(20).  if the first term is 5 and the 20th term is 45, then:\n\nS(20)=(10)(5+45)\nS(20)=500
130465
we need at least 12 terms\nSn = (n/2)*[2a1 + (n-1)d]\nin our series a1=11 and d=5. we need to find an n so that Sn>450:\n(n/2)*[2*11 + (n-1)5] = 450  /multiply equation by 2\nn*[22 + 5n - 5] = 900\nn*[5n + 17] = 900\n5n^2 + 17n - 900 = 0\nn1 = 11.82\nn2 = -6.92   /irrelevant\nwe want Sn to be greater then 450, and n has to be integer so the answer is n=12\nS12 = (12/2)*[2*11 + (12-1)*5]\nS12 = 6*[2*11 +11*5]\nS12 = 6*11*7 = 462 > 450
438145
Chisq test for proportional differences or a One-way Chi-sq same thing though.  The difference in your proportions +/- 1.96 x standard error.
644170
The answer of "nayanmange" is right but his formula is wrong because it does not predict the sequence correctly. The correct formula is (2n-1)-th and 2n-th term of the series are {(2n-1)^2+1} and {(2n)^2-1} respectively. For n = 3 we find the (2*3)-th i.e. 6th term of the series as {(2*3)^2-1} = 35.
254992
Well, you know that R=lim A_n /A_(n+1).\n\nNow do ratio tests on the problems:\n\nFor 2), convergence happens if lim |A_(n+1) z^3(n+1))/(A_n z^(3n)| <1. Thus, |z|^3 /R <1, so |z|<R^(1/3). This means the radius of convergence is R^(1/3).\n\nThe rest should be easy now.
313965
This approaches infinity as n approaches infinity.  To demonstrate, I'll show you something even smaller that sums to infinity.\n\n1 + 1/2 + 1/4 + 1/4 + 1/8 [4 times] + 1/16 [8 times]...\n\nThis is equal to 1 + 1/2 + 1/2 + 1/2 + 1/2.....which "goes to infinity.  \n\nSince each term my this series is equal to or smaller then your series (1/4 + 1/4 < 1/3 + 1/4, and 1/8 + 1/8 + 1/8 + 1/8 < 1/5 + 1/6 + 1/7 + 1/8, and so on), and my series approches infinity, your series must approach infinity, too!
52450
There is no answer.\n\nWhat you are expressing is the sum of ((-1)^n+1 * 2) from 0 to infinity. This is a divergent series and therefor has no answer, or in mathematica terms, the answer DNE (does not exist).