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bio website youtube.com/zonerarefatte
location San Diego, CA
age 25
visits member for 2 years, 4 months
seen Aug 21 at 14:59

Knowledge and understanding are quite different. Only understanding can lead to being, whereas knowledge is but a passing presence in it. (G.Gurdjieff)


Jun
22
awarded  Commentator
Jun
22
comment Closed form with of a series Mathematica
Yes, but what about the closed form if $x$ can be every value, so not necessarily $0<x<x_0$?
Jun
22
asked Closed form of a simple series
Jun
22
comment Closed form with of a series Mathematica
Ups, thank you so much!
Jun
22
revised Closed form with of a series Mathematica
added 172 characters in body
Jun
22
comment Closed form with of a series Mathematica
Added input code and yes $A=x_0$, $B=h$!
Jun
22
revised Closed form with of a series Mathematica
added 172 characters in body
Jun
22
comment Closed form with of a series Mathematica
Good idea! I didn't know about that!
Jun
22
awarded  Editor
Jun
22
revised Closed form with of a series Mathematica
added 200 characters in body
Jun
22
asked Closed form with of a series Mathematica
Jun
18
awarded  Scholar
Jun
18
accepted Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
Jun
18
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
yep, sorry. So it diverges. Thanks!
Jun
18
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
So, what is wrong in this proof of convergence? $$ S=\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_{0}|}=\frac{1}{x_{0}}\left(\sum_{k<x/‌​x_{0}}\frac{1}{x/x_{0}-k}+\sum_{k>x/x_{0}}\frac{1}{k-x/x_{0}}\right)=\frac{1}{x_{‌​0}}\left(\sum_{-k'<x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x_{0}}\frac{1}{k-x/x_{0‌​}}\right)=\frac{1}{x_{0}}\left(\sum_{k'>-x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x‌​_{0}}\frac{1}{k-x/x_{0}}\right)\leq\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{‌​x/x_{0}+k}+\frac{1}{k-x/x_{0}}\right) =\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{k^{2}-x^{2}/x_{0}^{2}}\right) $$
Jun
18
awarded  Student
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
This is my friend's proof of convergence: $$S=\sum_{k=-\infty}^{+\infty} \frac{1}{|x-kx_0|} = \frac{1}{x_0}\left(\sum_{k<x/x_0} \frac{1}{x/x_0-k}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right) = \frac{1}{x_0}\left(\sum_{-k'<x/x_0} \frac{1}{x/x_0+k'}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right)= \frac{1}{x_0}\left(\sum_{k'>-x/x_0} \frac{1}{x/x_0+k'}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right)\leq \frac{1}{x_0}\sum_{k>x/x_0}\left(\frac{1}{x/x_0+k}+\frac{1}{k-x/x_0}\right)= \frac{1}{x_0} \sum_{k>x/x_0} \left( \frac{1}{k^2-x^2/x_0^2} \right)$$
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
@Cocopuffs in this case the series is $\sum_{-\infty}^{\infty}$, not $\sum_{1}^{\infty}$!
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
Perfect, I arrive there. Then, how can I avoid the separation of the sum due to the study of the absolute value?
Jun
17
comment Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$
Perfectly captured. And thank you about this. So, the sum must diverge as one could aspect due to the similarity with the harmonic series? Actually, I'm absolutely with you on the error in my prof's proof, that I consider wrong.