Convergence of Euler's Infinite SeriesDate: 03/06/2002 at 21:00:17 From: Chris Subject: The Convergence of one of Euler's Infinite Series I was writing to inquire about the convergence of the infinite series 1+(1/2^2)+(1/3^2)+(1/4^2)+(1/5^2)+(1/6^2)+...+(1/n^2) = ? The hint given along with this problem was that the answer has something to do with the quadrature of a circle. I am a little bit stumped by this comment but believe it has something to do with pi. My first attempt at solving this problem was to try to split it between the even and odd denominators. So [(1/2^2)+(1/4^2)+(1/6^2)+...] + [1+(1/3^2)+(1/5^2)+(1/7^2)...] From here i attempted to find the sum of each part, but both have escaped me. My second attempt was to approximate the decimal value and work backward. Using a computer program I approximated the decimal value to 1.6449, but this is not a recognizable fraction and it does not seem to be growing fast enough to approach 1.666666666 Any help in solving this series would be greatly appreciated, preferably an answer and perhaps some general advice when solving infinite series. Thank you in advance for your time! Chris Date: 03/07/2002 at 00:52:17 From: Doctor Schwa Subject: Re: The Convergence of one of Euler's Infinite Series Hi Chris, The only solutions I know to this sum uses quite a bit of calculus. One nice method is in our archives at Sum of 1/n^2 http://mathforum.org/dr.math/problems/schmidt.7.24.00.html and my favorite method is at Euler's summmation of 1/n^2 http://mathforum.org/dr.math/problems/bennett.2.15.00.html and yet another method is at Infinite Series Involving Pi http://mathforum.org/dr.math/problems/tak12.16.97.html Give those a try, and feel free to write back if you still have questions. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 03/07/2002 at 19:10:31 From: Chris Subject: Re: The Convergence of one of Euler's Infinite Series Dear Dr. Schwa, Thank you many many times for your help with that problem. Sadly, I don't fully grasp the solutions you presented and I am afraid the problem is on my end. To aid in my understanding, do you think you could explain the "infinite polynomial product" theorem and the "power series expansion" you used in your solution of the problem (the second link wherein you talked about the function f(x) = sin (pix)? Neither my previous math courses nor my current calculus class have introduced these concepts yet. Thank you in advance once again! Chris Date: 03/07/2002 at 19:31:25 From: Doctor Schwa Subject: Re: The Convergence of one of Euler's Infinite Series Hi Chris, The infinite polynomial part is within your reach from only an algebra class. sin(pi x) is zero for all integers x. Thus, just as a polynomial with k as a zero has to have (1 - x/k) as a factor, so must this function. Perhaps you're more used to seeing it as (x-k), but in this case that would create a function that blows up to infinity. Dividing through by k maintains the zero and gives a finite function that you can then set to the right height by matching the slope at x = 0 (which is pi). The power series expansion comes toward the end of calculus BC, but the idea is a generalization of the linear approximation. That is, sin (x) is approximately sin(0) + x * (the derivative of sin at 0), or more generally, f(x) = about f(0) + x * f'(0). That just means to follow the tangent line. You can generalize this, to follow the parabola, or best cubic, or so on, and find f(x) = approximately f(0) + x * f'(0) + (x^2 / 2) f'''(0) + (x^3 / 6) f'''(0) + ... from which you get sin(pi x) = approximately 0 + pi x * 1 + 0 - (x^3 / 6) pi^3 + ... Then you can set the coefficients of x, and of x^3, equal to each other in these two different methods ... Does that help clear things up? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 03/08/2002 at 18:18:58 From: Chris Subject: Re: The Convergence of one of Euler's Infinite Series Dear Dr. Schwa, Thank you very much for the clarification; that makes the solution infinitely (no pun intended) clearer. That pretty much answers my questions. Thank you very much for your time; I think you and all the Dr. Math people are doing a great service by helping to stimulate the desire to do mathematics. Thanks again! Chris |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/