- #1
Yuqing
- 218
- 0
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.
[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.
Last edited: