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        riddles >> putnam exam (pure math) >> Convergent Series
        
(Message started by: ThudanBlunder on Apr 1^st, 2007, 5:20pm)

Title: Convergent Series
Post by ThudanBlunder on Apr 1^st, 2007, 5:20pm

For what range of a > 0 does the series below converge?

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifa^{(1 + 1/2 + 1/3 + ....... + 1/n)}
n=1

Title: Re: Convergent Series
Post by kris on Apr 1^st, 2007, 8:42pm

My answer : [hide]a < 1/e [/hide].

Title: Re: Convergent Series
Post by Icarus on Apr 2^nd, 2007, 5:05pm

Providing details: Let L(n) = 1 + 1/2 + ... +1/n. Obviously, the series diverges for a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1, so assume a < 1.

ln(n) < L(n) < ln(n) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif, so a^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subgamma.gif}http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif a^ln(n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif a^L(n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif a^ln(n). Hence the series converges or diverges with http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif a^ln(n). But a^ln(n) = n^ln(a), and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif n^ln(a) is well known to converge if and only if ln(a) < -1.

Hence http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif a^L(n) converges if and only if 0 < a < 1/e.