Sum of 1/Sqrt(i)Date: 11/20/2000 at 13:11:23 From: Jeff Khoury Subject: Mathematical induction, sum What's the formula for n SUM [1/sqrt(i)] i=1 This question was asked in a college quiz. Date: 11/20/2000 at 16:21:17 From: Doctor Rob Subject: Re: Mathematical induction, sum Thanks for writing to Ask Dr. Math, Jeff. I don't know a formula for the sum. I do know that the sum lies between 2*sqrt(n)-2 and 2*sqrt(n). That is because what you have is bounded above by INT [1/sqrt(t) dt] from 0 to n and below by INT [1/sqrt(t) dt] from 1 to n which have the values I indicated. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 11/21/2000 at 11:59:32 From: Jeff Subject: Re: Mathematical induction, sum Sorry to bother you again, but if my math teacher asked for a formula then it should exist. How come you can't find it? Thank you. Date: 11/21/2000 at 12:57:55 From: Doctor Rob Subject: Re: Mathematical induction, sum Thanks for writing again. I don't have a good answer to your question. I have thought about this for two days, and asked several very knowledgeable friends. We all believe that no such formula exists. If your instructor has such a formula, we would be very interested in seeing it. I do know that the above sum diverges as n increases without bound, because of the limits I quoted above. In fact, my friends and I believe that n lim [SUM 1/sqrt(i) - 2*sqrt(n)] n->infinity i=1 is a well-known quantity, zeta(1/2) = -1.4603545..., which is believed to be a transcendental number. Of course this limit doesn't give you a formula for the sum. Sorry! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 11/21/2000 at 15:54:56 From: Doctor Rob Subject: Re: Mathematical induction, sum Dear Jeff, One of my friends has devised a formula for your sum. It is not simple, but here it is: n SUM k^(-1/2) = 2*n^(1/2) - 3/2 + n^(-1/2)/2 - k=1 n (1/2)*INT (x-[x]-1/2)*x^(-3/2) dx) 1 Here [x] means the greatest integer less than x. This seems very likely to be amenable to proof by induction. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 11/22/2000 at 08:49:55 From: Jeff Subject: Re: Mathematical induction, sum Dear Dr. Rob, I have to tried to prove the formula you gave me by induction and failed. I'm asking you (I hope I'm not asking too much) to send me the proof of this formula. Thank you a lot for your help. Date: 11/27/2000 at 08:58:44 From: Doctor Rob Subject: Re: Mathematical induction, sum Thanks for writing back, Jeff. First you need the basis of the induction. When n = 1, both sides are equal to zero, because the integral is from 1 to 1, and hence has value zero. Now assume it is true for n, and try to prove it for n+1. Then starting with the right side for n+1, you have 2*(n+1)^(1/2) - 3/2 + (n+1)^(-1/2)/2 - n+1 (1/2)*INT [x-[x]-1/2)*x^(-3/2) dx 1 = 2*(n+1)^(1/2) - 3/2 + (n+1)^(-1/2)/2 - n n+1 (1/2)*INT f(x) dx + (1/2)*INT f(x) dx 1 n Now use the induction hypothesis to replace the first integral by its equal from the statement for n, which you are assuming is true. The last part is to evaluate the integral n+1 INT (x-[x]-1/2)*x^(-3/2) dx n This is made easier since [x] = n in this interval of integration, so you get n+1 INT x^(-1/2) - n*x^(-3/2) - (1/2)*x^(-3/2) dx n which involves only integrating powers of x. When you do this, you should get terms which cancel with the ones in the rest of the equation, and you should be left with n n+1 SUM k^(-1/2) + (n+1)^(-1/2) = SUM k^(-1/2) k=1 k=1 which is the left-hand side of the statement for n+1. You can see that this proves that the right-hand and left-hand sides of the statement for n+1 are equal, assuming they are for n, so the induction step is complete. Now invoke the Principle of Mathematical Induction to conclude that the statement is true for all natural numbers n. You fill in the details. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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