Let {x_n} be a sequence in a metric space .Prove {x_n} converges iff every proper subsequence of {x_n} converges
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A => B\nBy reductio ad absurdum: suppose there exists a proper subsequence {x_k(m)} diverging. This means that, for any given point X of the metric space and for any given distance M, there exists m such that d(X, x_k(m)} > M.\nDefine n=k(m) and you have that the main sequence would also diverge.\n\nB=>A\nAgain by reductio ad absurdum: suppose the main sequence {x_n} diverges. This means that for any X and for any M, there is n such that d(X, x_n} > M.\nNow take 2M, then 3M, 4M, etc. You have built a function n=k(m) corresponding to a diverging subsequence {x_k(m)}.
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The definition of a limit is slightly different depending on whether you are looking at the limit of a function at some point&#92;, or the limit of a function at infinity, or the limit of a sequence at infinity. I will do the last.\n\nWe say that the limit of the sequence a_n as n goes to infinity is the number L if whenever e>0, there is a natrual number N so that whenever n>=N, we have |a_n -L|<e.\n\nIn other words, the limit is L if for every 'tolerance' e, we can go far enough out in the sequence (N) so that every term of the sequence farther out is within the tolerance of the purported limit.\n\nAs an example. Let a_n=1/2 +1/4 +...1/2^n.\nIt is easy to show by induction that a_n =1 - 1/2^n.\n\nGiven a tolerance e>0, choose N so that 1/e <2^N. Then, if n>=N,\nwe have |a_n -1|=1/2^n < e.\n\nThis shows that the limit L is 1. It has to be proved that limits are unique when they exist, but that is not too difficult. Any good advanced calculus course will cover this stuff.
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you have to do it by induction:\nbasically:\nn=2:\nka_1 + ka_2 = k( a_1 +a_2)\n\nnow, assume it is true for n\n\nlet us prove it for n+1:\n\nsum_{i=1} ^{n+1} ka_i = sum_{i=1} ^{n} ka_i + ka_{n+1}\n= k ( sum_{i=1} ^{n} a_i )+ ka_{n+1} now you have only two summands, so:\n= k( sum_{i=1} ^{n} a_i  + a_{n+1} )\n= k sum_{i=1} ^{n+1}  a_i \n~
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The sum from n = 0 to infinity of a^n is called a geometric series. It converges if and only if |a| < 1.\n\nIn this case we have the sum from n = 1 to infinity of (x^n)/(9^n), which can also be written as (x/9)^n. So in this case a = x/9, and in order to converge we need |x/9|<1. Thus the interval of convergence is -9 < x < 9.\n\nAs for the sum of the series, well the sum from n = 1 to infinity of a^n is a/(1-a). Here a = x/9, so you've got (x/9)/(1-x/9) = x/(9-x).\n\nBTW, the series is still geometric even if x is not a natural number. Also x needs to be greater than -9.
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thats a geometric sequence with a common ratio of -1... you can get the common ratio by getting the first set of differences... thats 10-12... 12-11.. 11-15.. 15-12... then have the first set of difference: -2 1 -4 3.. then get the second set of difference... -2-1.. 1--4... -4-3... and u get -1 -1.. thats how u get to chek if its a geometric sequence by checking if they have the same second diff... chek the net and search for the formula for solving geometric sequences cuz i forgot bout those already.. goodluck!
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The full version of Tychonoff's involving uncountable products doesn't come up very often in Algebraic Topology. Of course, the result for finite products is pervasive.\n\nIn functional analysis, the first big application is in the proof of Alaoglu's theorem about the weak* compactness of the unit ball of the dual of a Banach space. This result has several generalizations to other topological vector spaces such as the weak operator topology for the space of bounded operators on  a Hilbert space. It is used to show various representation results for Banach spaces including that every Banach space can be embedded as a closed subspace of some C(K) space (where K is compact Hausdorff). The same method of proof comes up when considering the Gelfand structure space for commutative Banach algebras.
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That uses 'x' a couple of times, so just to make things clearer, I'll call the point we're considering 'y'.\n\nLets call the limits of f and g at y A and B respectively. \nNow, by definition of a limit: \nFor all ε > 0, there exists δ1 > 0 such that for all x with 0<|x-y|<δ1, |f(x)-A|<ε.\nSimilarly,\nFor all ε > 0, there exists δ2 > 0 such that for all x with 0<|x-y|<δ2, |g(x)-B|<ε.\n\nNow, lets suppose (for the sake of contradiction) that B<A. Lets make ε = (A-B)/2. Then there exist δ1 and δ2 such that the above holds.\nNow, from above there exists δ1 > 0 so that if |x-y|<δ1 then |f(x)-A|<ε, and δ2 > 0 so that if |x-y|<δ2 then |g(x)-B|<ε.\n\nLets pick x close enough to y so that both those conditions hold - eg x = y+min(δ1,δ2). Then the following holds:\ng(x) < B+ε = B+(A-B)/2 = A-(A-B)/2 = A-ε < f(x). Contradiction.\n\nThe key idea is that if B>A, then there is a 'gap' between B and A - thus you just need to get smaller than the gap, and you'll get a contradiction.
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35\n\nThe sequence is every even-indexed Fibbonacci number plus 1.  So from the original sequence whose value is the sum of the two previous numbers:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...\n\nYou take every second number and add 1:\n\n2, 3, 6, 14, 35, 90, 234, ...
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I apologize if this is too simple, but it seems this is the obvious answer:\n\nSU(2) is a collection of 2x2 matrices, so the sequence { s_j } is just a subset.  The diagonal matrices in the special unitary group consist of [[a,0],[0,conj a]] where |a| = 1.  A "diagonal subsequence" should be a subset of this group.  You should be able to identify such a subsequence with complex numbers which lie on the unit circle.
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I'm not familiar with this sequence.  \nHave you tried googling it?  Make sure you use quotes.\n"perrin's sequence"\n\nHere's what I have found:\nDefine the sequence of integers by the linear recurrence formula\n               s[n] = s[n-2] + s[n-3]\nwith the initial values  s[0] = 3, s[1] = 0 , s[2] = 2. \nBased on the above values,\ns(3) = s(3-2) + s(3-3)\n= s(1) + s(0)\n= 0+3\n=3\ns(4) = s(4-2) + s(4-3)\n= s(2) + s(1)\n= 2+0 = 2\ns(5) = s(3) + s(2)\n= 3+2 = 5\nand so on...
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i think its true\nim f is isomorphical  with G/(ker f) so\n|G| = |ker f|*|G/(ker f )| = |ker f| *|im f|.\n i applied two well known theorems.
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The previous result is that the number of elements in the preimage of each element in f(K) (or more generally in f(G)) is n. All those sets are disjoint and their union is K, so the total number of elements in K is n * (Number of elements in f(K)).\n --> |K| is a multiple of |f(K)|\n\nAnother way of looking at is that if N is the kernel of f, f(K) is isomorphic to the quotient group K/N which always divides the order of K.
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if B' is subset of A' then\n\nif x belongs to B' then x belongs to A'\n\nif x belongs to A then x cannot belong to B' because if x belongs to B' then it belongs to A then x belongs to A and A'\n\nso if  x belongs to A the x belongs to B(as it does not belong to B')\n\nso A is subset of B
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#1: Yes, that is correct.\n#2: This series diverges. It is a necessary (but not sufficient) condition for an infinite series to converge that the individual terms approach 0 as n→∞. It is fairly easy to show that the individual terms approach 1 as n→∞.\n#3: First note that ln ((n+1)/n) = ln (n+1) - ln n. So writing out the first few terms of this series:\n\n(ln 2 - ln 1) + (ln 3 - ln 2) + (ln 4 - ln 3) + ...\n\nNote that the second part of each term cancels out the first part of the preceding term. So all that needs to be considered is the first part of the last term and the last part of the first term, because those are the only parts that do not cancel. Formally:\n[n=1, a]∑ln ((n+1)/n) = ln (a+1) - ln 1 = ln (a+1)\n\nThis series does not converge, since [a→∞]lim ln (a+1) = ∞.
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well the series is only convergent if the partial sums approach zero as n approaches infinity\n\nn^2(e^(-n^3) = n^2/(e^(n^3)\n\nnow take  lim n->inf n^2/(e^(n^3)\n\ne is being rasied to a number that gets very large very quickly as opposed to n squared so this should approach zero\n\nto prove that we can just use L'Hopistal's rule once and see that this then becomes lim n->inf 2/(3ne^(n^3)) this is obviously zero\ntherefore since the partial somes become zero eventually.. the total series converges ... you continue to add zero to the total sum as n goes to infinity\n\nrusty on my calc two hopefuly that helps
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You can start from the following identity for set differences which holds true for any sets X, A, and B\n\nX − (A ∪ B)  =  (X − A) ∩ (X − B)\n\nNow, if B is a subset of A, then A ∪ B = A\n\nthe identity above becomes\n\nX − A =  (X − A) ∩ (X − B)\n\nFor this to hold true, X − B must be a superset of (X − A), or in other words, (X − A) must be a subset of X − B
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Add some rigor. Remember that a sequence {a_i} is defind to converge to a limit L iff for every positive difference D > 0, there exists some n such that for all N > n, the difference between the limit L and any subsequent term a_N is less than D. We are squeezing D down to 0 by showing that we can always supply such an n. Thus, your job is to supply some pseudofunction n(D) giving an upper range of indices that satisfy the definition above.\n\nFor the pattern above, the (2n + 1)th term is 8 + 1/(n + 2) while the (2n)th term is 8 + 2/(n + 2). As n increases without bound, the (2n + 1) subsequence approaches 8, as given any positive difference D, one can find a least (2n + 1) such that all terms greater than (2n + 1) yields a difference 8 - (8 + 1/(n + 2)) = 1/(n + 2) less than D. In this case, choose n(D) > (1 - 2D)/D. \nUse a similar tactic to show that the even indexed subsequence converges to the same value, and thus the sequence converges to 8.\n\nAs for the second sequence, as long as all of the remaining values are 6, it converges to 6 by a trivial argument. Given a positive difference D, simply choose n(D) > 6, since 6 - a_n = 0 for any n > 6.