Alternating Harmonic Series
in [Math]
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From: Maury Barbato on 9 Apr 2010 03:03 Hello, consider the alternating harmonic series, that is the series sum_{n=1 to infty} [(-1)^(n+1)]/n, and let sum_{n=1 to infty} a_n be the Cauchy product of the alternating harmonic series with itself. (I) Is {|a_n|} a decresing sequence? (II) Is sum_{n=1 to infty} a_n convergent? Thank you so much for your attention. My Best Regards, Maury Barbato
From: David W. Cantrell on 9 Apr 2010 08:13 Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > Hello, > consider the alternating harmonic series, that is the > series sum_{n=1 to infty} [(-1)^(n+1)]/n, and let > sum_{n=1 to infty} a_n be the Cauchy product of the > alternating harmonic series with itself. > > (I) Is {|a_n|} a decresing sequence? Yes. In fact, a_n = (-1)^n 2(gamma + psi(n))/n where gamma is the Euler-Mascheroni constant and psi denotes the digamma function. > (II) Is sum_{n=1 to infty} a_n convergent? Yes. It converges to (log(2))^2. David
From: Maury Barbato on 9 Apr 2010 05:26 David W. Cantrell wrote: > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > Hello, > > consider the alternating harmonic series, that is > the > > series sum_{n=1 to infty} [(-1)^(n+1)]/n, and let > > sum_{n=1 to infty} a_n be the Cauchy product of the > > alternating harmonic series with itself. > > > > (I) Is {|a_n|} a decresing sequence? > > Yes. In fact, a_n = (-1)^n 2(gamma + psi(n))/n > > where gamma is the Euler-Mascheroni constant > and psi denotes the digamma function. > > > (II) Is sum_{n=1 to infty} a_n convergent? > > Yes. It converges to (log(2))^2. > > David Thank you David for your help: I was going crazy trying to study this series! Anyhow, I have just found a proof at p. 84 of the wonderful book "An Introduction to the Theory of Infinite series" by Bromwhich. My Best regards, Maury Barbato
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