
#1
Feb1811, 01:19 PM

P: 218

I was reading a book which is a collection of interesting mathematical journal articles. Within the book there was an article which discussed alternating series. In particular, at one point in the article it proves that the series
[tex]\sum (1)^n \frac{\sqrt(2)^{(1)^{n}}}{n}[/tex] diverges. To be a bit more clear the series is [tex]\frac{\sqrt{2}}{1}\frac{1}{2\sqrt{2}}+\frac{\sqrt{2}}{3}\frac{1}{4\sqrt{2}}+..[/tex] To directly quote the article: "Since [tex]lim (1)^n \frac{\sqrt(2)^{(1)^{n}}}{n} = 0 [/tex] we may group these terms in pairs, and number them 2n+1, 2n+2 in pairs, with n=0,1,2..., as follows: [tex]S=[\frac{\sqrt{2}}{1}\frac{\sqrt{2}}{4}] + [\frac{\sqrt{2}}{3}\frac{\sqrt{2}}{8}] + ...+ [\frac{\sqrt{2}}{2n+1}\frac{\sqrt{2}}{4n+4}][/tex] [tex]S=\sum [\frac{\sqrt{2}}{2n+1}\frac{\sqrt{2}}{4n+4}][/tex] [tex]=\sum \frac{\sqrt{2}}{4} \frac{1}{n+1} \frac{2n+3}{2n+1}[/tex] where the latter part is clearly divergent. " Would that be considered a rearrangement of the series? I'm a bit confused on whether you can group terms together or move them around. I know series which are conditionally convergent can be rearranged to any sum and it follows that you cannot prove a limit with a rearrangement. To illustrate the source of my confusion I look at the classic series 1+(1)+1+(1)+1+(1)+... in which 1+(1+1)+(1+1)+... gives 1 and (1+(1))+(1+(1))+(1+(1))+... gives 0 so grouping does seem to affect the result and this journal appears to be doing the same thing. Also, I'm not sure what the significance of the first statement (the limit > 0) is. It doesn't seem to be used. 



#2
Feb1811, 01:26 PM

HW Helper
P: 2,150

By rearrangement we mean infinite rearrangement.
What was done here is even partial sums have been considered. Clearly if the even, odd, or any infinite subset of partial sums diverge the series diverges. The the limit > 0 is used in that it is a necessary condition for the series to converge, we can say the original series converges if and only if the new one does. It is not needed in that the series diverges, but if the new series had converged the limit condition would be needed to conclude the convergence. It is in a sense easier to show a series diverges because many of our theorems and lemmas are of the form if several things are true the series converges, if any of them are false the series diverges. To prove convergence we need to check all the conditions to prove divergence we can stop when any condition fails to be satisfied. 



#3
Feb1811, 06:02 PM

P: 76





#4
Feb1811, 08:01 PM

HW Helper
P: 2,150

Rearranging a series to prove a limit.
^I clarified above I was speaking of the partial sums not terms of the series. The sequence {S_N} dervived from a sequence {a_N} by the rule S_(N+1)=S_N+a_(n+1).




#5
Feb1811, 09:14 PM

P: 76




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