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Targil 13 – limits. sin m t dt , where m and n are natural numbers, 1. Compute lim x 0 tn x (the answer may depend on m and n). 2x Solution. By mean value theorem of integration 2x sin m t sin t 1 sin y dt dt n n m x t x t t y m 2x m 2x t 1 nm dt , where x ≤ y ≤ 2x. x sin y sin y 1 1 . Therefore , so x 0 x 0 y y 2x 2x 2x sin m t 1 mn 1 1 lim dt lim nm dt nm1 lim m n nm1 1 nm1 n x0 x0 x0 t t t 2 x x x x if n m 1. Since the integral is positive anyway, it only matters whether 1 lim nm1 tends to zero or to infinity in these cases, so x 0 x if n m 1 then it tends to infinity, if n m 1 then it tends to zero, and if n m 1then 2x 2x sin m t 1 lim dt lim dt lim ln 2 x ln x ln 2 n x 0 x 0 x 0 t t x x Remark. This problem is pretty obvious, but it really did appear on IMC. m Hence 2. a) Does 1 k ln k ln ln k converge? k 100 b) Does k k 100 1 ln k ln ln k 2 converge? Solution. a) Apply the “harmonic trick” : split the sequence into sub-strips: strip number n will contain numbers between 2n+1 and 2n+1. Then all numbers in the strip number n are (up to a bounded factor) 1 and there are 2n elements in this strip. n 2 n ln n Therefore the sum of that strip is 1 . n ln n So it is enough to investigate the convergence of the series 1 n ln n . n ? Apply the harmonic trick again: cut the sequence into strips, strip number m will contain n’s between 2m+1 and 2m+1. 1 The elements in strip number m are, up to a bounded factor m , this strip 2 m 1 has 2m elements, so the sum in the strip is, up to a bounded factor . m 1 So the sum over all strips behaves like , and that one diverges (the m ? m easiest proof for that is applying harmonic trick again. b) As in a), apply the harmonic trick twice. Each time you get the series with 1 the same convergence properties. After firs time you get , and 2 n ln n n ? after second time you get 1 m m ? 2 , and that one converges. 3. Let {an} be a sequence defined by a0 = 1 and recursive formula 1 n ak an1 n 1 k 0 n k 2 a Find the limit kk . k 0 2 Solution. Consider a generating function of that sequence f x an x n . n 0 1 Then the limit of the series that we should compute is simply f . 2 n ak The equation n 1 an1 after we multiply both sides by xn and k 0 n k 2 n x nk xm k ak x k sum them up turn to n 1 an1 ak x n k 2 m 0 m 2 k 0 n 0 n 0 k 0 xm m 0 m 2 f ' x f x xm 1 x x 2 x3 ... . 2 3 4 5 m 0 m 2 x 2 x3 x 4 x5 But we know that ln 1 x x ... 2 3 4 5 x ln 1 x Therefore g x . x2 x ln 1 x df dx Hence f x2 So, we should try to compute g x ln 1 x ln 1 x dx 1 ln f const dx ln x x x 1 x dx x x2 \ ln x ln 1 x ln 1 x 1 1 1 dx ln x dx x x 1 x x x 1 x ln 1 x 1 x ln x ln 1 x ln 1 x x x When x = 0 the right hand side might be seen as a limit (since it comes from power series) so it is –1 ; f (0) = 1 and its ln is 0, hence left hand side is equal to const, so const = –1. 1 x ln 1 x So f x exp 1 x e 1 1 Therefore f exp 1 ln e1ln 2 2 2 2 ln x 4*. Compute lim sin tan x tan sin x . arcsin arctan x arctan arcsin x Simple lemma. Suppose we have a function f (x) such that f (0) = 0 and f has a continuous derivative around 0. f u f u Suppose also that u, v 0 and always u v . Then f ' 0 . u v f u f u Proof. By Lagrange theorem, f ' w where w is between u u v and v, so when u, v 0 , w also tends to 0. QED of lemma. x 0 Notice that all functions are analytic. sin tan x tan sin x at least sometimes, for example when 3 x arctan because, then LHS is negative and RHS is positive. 2 Since functions are analytic, their points of coincidence are isolated, so at some neighborhood of 0, except for 0 itself, sin tan x tan sin x , so we learn from complex functions. So, at some neighborhood of 0 their inverse functions, arcsin arctan x ,arctan arcsin x also don’t coincide, hence we can consider that fraction without problems. From now on we shall say p(x) ~ q(x) to indicate that lim x 0 p x 1. q x It is known that x ~ sin x ~ tan x. Denote y tan sin x then y ~ x. lim x 0 sin tan x tan sin x arcsin arctan x arctan arcsin x lim x 0 lim x 0 sin tan arcsin arctan y y arcsin arctan x arctan arcsin x x tan sin arctan arcsin x sin tan arcsin arctan y y The last equality follows from the above lemma, when you apply function f x tan sin x to u arcsin arctan x , v arctan arcsin x , you get If we denote g y sin tan arcsin arctan y , and its inverse function g x tan sin arctan arcsin x , then what we got is arcsin arctan x arctan arcsin x ~ x tan sin arctan arcsin x . 1 lim x 0 g y y x g 1 x From x ~ sin x ~ tan x it follows that x ~ g x ~ g 1 x . But we have seen before that both numerator and denominator are nonzero, so g has no stable points in the neighborhood of 0 other than 0. Hence Taylor series of g has at least 2 nonzero terms, and the first is x: g x x ax n ... , where a is an nonzero number which I know nothing about, and n > 1 is a number which I know almost nothing about, and I don’t want to (though they can be computed). Then g 1 x x ax n ... . Now since y ~ x, ay n o y n g y y lim lim 1 x 0 x g 1 x x 0 ay n o y n So, the answer is 1. 5*. {an} is a sequence of positive real numbers such that an + am ≥ an+m for all m, n. Prove that the sequence an/n converges. Proof. Firstly, an ≤ na1 hence the sequence an/n is bounded from above. It is also bounded from below by 0. So it is bounded. Hence it has liminf (lower limit) which will be denoted by L. For each ε > 0, we can find an index k such that ak/k < L + ε. Then ak < k(L + ε) . Then amk ≤ mak < mk(L + ε) . For each natural N, we can write N = mk + r , where r < k (division with remainder). Hence aN ≤ amk + ar ≤ mak + ra1 < mk(L + ε) + ma1 . Therefore aN / N < L + ε + ma1 / N < L + 2ε , for sufficiently large N. So, for any ε > 0, aN / N < L + 2ε, for sufficiently large N. Since L is liminf, then for sufficiently large N also aN / N > L – 2ε . Hence the sequence converges to L.