Tricks to Sum an Infinite SeriesDate: 01/22/99 at 22:38:20 From: Steve Heggie Subject: Convergence of a strange sum infinity SUM [((-1)^n)*|sin n|]/n n = 1 Do you have any ideas on how to find the convergence of this series? I'm pretty sure it converges on zero, but I cannot begin to prove it (aside from just adding up the partial sums). Thanks, Steve Date: 01/23/99 at 08:19:13 From: Doctor Anthony Subject: Re: Convergence of a strange sum The series is -sin(1) + sin(2)/2 - sin(3)/3 + ....... = -IM[e^(i) - e^(2i)/2 + e^(3i)/3 - ..... where IM means 'imaginary part of'. Let e^i = x and the series can be written: = -IM[x - x^2/2 + x^3/3 - x^4/4 + ....... Differentiating the series we get: = -IM[1 - x + x^2 - x^3 + ....... = -IM[1/(1+x)] To find the sum of the actual series we must integrate this, to get: = -IM[ln(1+x)] = -IM[ln(1+e^i)] Suppose ln(1+e^i) = p+iq and we require the value of -q. Then: 1 + e^i = e^(p+iq) 1 + e^i = e^p e^(iq) 1 + cos(1) + isin(1) = e^p[cos(q) + isin(q)] 1 + cos(1) + isin(1) = e^p cos(q) + i e^p sin(q) Therefore: 1 + cos(1) = e^p cos(q) sin(1) = e^p sin(q) Dividing these equations: tan(q) = sin(1)/[1+cos(1)] q = tan^(-1)[sin(1)/(1+cos(1))] = 0.5 We require -q as the sum of the given series, so the required sum is -1/2. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 01/23/99 at 16:23:39 From: Steven Heggie Subject: Re: Convergence of a strange sum Dr. Anthony, I've been working through the solution that you provided to my infinite sum problem ((-1)^n * |sin (n)|) / n, and I have figured everything out except for one part. It is the line that suggests that i sin(1) = i e^p sin(q). I don't understand why. Also, can you explain the line right after that? Thank you, Steve Date: 01/23/99 at 17:49:55 From: Doctor Anthony Subject: Re: Convergence of a strange sum When you have a complex equation like: a + ib = p + iq then we can equate the real part on the left with the real part on the right, that is, a = p, and equate the imaginary part on the left with the imaginary part on the right, that is, b = q. It is easy to prove this. Write the equation: a-p = i(q-b) Now square both sides: (a-p)^2 = -(q-b)^2 since i^2 = -1 (a-p)^2 + (q-b)^2 = 0 This requires the sum of two positive terms to equal 0. Clearly this is impossible and in fact each bracket must separately equal 0. That is, a = p and b = q. Applying this idea to the equation: 1 + cos(1)+i sin(1) = e^p cos(q) + i e^p sin(q) We must have: 1 + cos(1) = e^p cos(q) equating real part on each side sin(1) = e^p sin(q) equating imaginary part on each side I hope that has cleared up the difficulty. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 01/23/99 at 18:02:35 From: Steven Heggie Subject: Re: Convergence of a strange sum Dr. Anthony, Thanks for the help. I NEVER would have figured this problem out on my own. Thank goodness for Euler's Formula! - Steve Heggie |
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