A1=5     A2=(4n+1)/(5n+3)  Determine whether the sum of A sub n converges?
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.
766268
This series converges to \n(sqrt(3)*pi/4)-(4ln(2))+\n((9/4)ln(3)).
680780
You can use infinite series...\n0 = 0+0+0+...\nsubstitute 0 = (1-1)\nso \n0 = (1 - 1) + (1 - 1) + (1 - 1) + ...\n\nuse associatve law of addition so that\n0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... \n\nthen cancel out leaving you with \n0=1\n\nerror: using associative law on on infinite sum that does not converge
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
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!
335533
Well, you know C<4 to start with.\n\nAlso, since the total sum is odd, either B or C must be odd, but not both.  Since C has the fewest possible values, I'd start with that.\n\nIf C=3, 5C = 15 and there is no combination of 2A + 3B (where A and B must be 1 or more) that could equal 4.  So C must be 1 or 2.\n\nIf C=2, 2A + 3B must equal 9.  So B must be 3 or less.  If B = 3, A = 0, which is not acceptable.  B must be odd (see above paragraph), so B must be 1 (yielding A = 3).\n\nIf C = 1, 2A + 3B must equal 14.  So B must be 1, 2, 3 or 4.  As in the paragraph above, if C is odd, B must be even.  So if B = 4, 2A + 12 + 5 = 19, so A = 1.  If B = 2, 2A + 6 + 5 = 19, so A = 4.\n\nWe now have 3 solutions:\nI) A = 3, B = 1, C = 2\nII) A = 1, B = 4, C = 1\nIII) A = 4, B = 2, C = 1
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
1139003
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
632401
u(n)=1/3(1+5(-2)^(n))\n\nConsider the even values of n; i.e. 0,2,4,6...\nu(0)=2, u(2)=7, u(4)=27... it will diverge.\n\nConsider the odd values of n; i.e. 1,3,5,7...\nu(1)=-3, u(3)=-13, u(5)=-53... it will also diverge.\n\nAs the numerical values of both the odd and even values of n are getting bigger as n increases, the function, u(n) must diverge.
203992
a) [1,2] is a sub set of domain of g so for every x in [1,2]\nwe have a unique value of g.\n\nb) if the sequence converge, in the limit Xn+1 must be equal to X, so if the limit is L we have:\n\nL=(3L^2+3)^1/4\n\nL^4=3L^2+3\n\nL^4-3L^2-3=0\n\ndelta=9+12=21\n\nL^2=(3+√21)/2  (L^2>0)\n\nL = √[(3+√21)/2]  (L>0 because X0>0)\n\n4<√21<5  ---> 7<3+√21<8\n\n---> 3<(3+√21)/2<4  ---> 1<L<2\n\nc) we have (Xn+1)^4=3Xn^2+3 let Yn be Xn^2\n\n---> (Yn+1)^2=3Yn+3 ---> Yn
925167
Hi Karen \n\n5(w+3)+3(w+1)=14\n\n● Step 1\n5(w+3) = \n(5 * w ) + ( 5 * 3 ) =\n5w + 15\n\n● Step 2\n3(w+1) =\n( 3 * w ) + (3 * 1) =\n3w +3 \n\n● Step 3\n5w + 15 + 3w +3  = 14\n( 5w + 3w ) = 14 - 15 - 3\n8w = -1 -3 \n8w = -4\nw = - 4 / 8\nw = -1/2  Or  - 0.5 \n\nGood Luck dear ♣
598513
n + (n+1) + (n+2) = 102\n3n + 3 = 102\n3n = 99\nn = 33\nn+1 = 34\nn+2 = 35\nsecond integer = 34
992739
You mean the sum of the first integers, starting with 2?\nWrite 2 in A1\nWrite 5 in A2\nWrite =2*A2-A1+1 in A3\nWrite =2*A3-A2+1 in A4\nWrite =2*A4-A3+1 in A5\nand so on. It can be continued by dragging.\n\nIf you only want the sum of 2 to N you can use\n= 0.5*N*N+1.5*N
85542
A(5) = 1/2 + (5-1)(1/4)\nA(5) = 1/2 + (4)(1/4)\nA(5) = 1/2 + 1\nA(5) = 1 1/2
857511
N=1/4+3/5\n(5+12)/20\n=17/20
566299
(n/n-1)(1/n)(n/n+1)=\n\n[n^2]/[(n-1)(n+1)n]=\n\n[n]/[n^2-1]=5/k\n  \n\nso, k=5*[(n^2-1)/n]\n\nso, if for example n=5 then k=24.
609611
Dear amit.wizkid, thou little knoweth what an abysmal error thou maketh.\n\ntut_einstein, you are better off noting that the sum of the first 'n' terms of an AP, whose first term is 'a' and the difference between terms is 'd' is :\n\nS = (n/2)[2a + (n-1)d]\n\nYou state that \nS1/S2 = (n+1)/(n+3)\n\nwhich is \n\n[2a1 + (n-1)d1] :[2a2 + (n-1)d2] = (n+1) : (n+3) ..... (I)\n\nRemember that what you desire to know is\n(a1 + 5d1) : (a2 + 5d2) \n\nPut n = 11 in (I)\n\n[2a1 + 10 d1] :[2a2 + 10 d2] = 12:14\n\nwhich gives\n\n[a1 + 5d1] : [a2 + 5d2] = 6:7\n\nwhich is what you desire. \n\nShould suffice.