N nice rose Nov 2009 23 0 May 18, 2010 #1 how to find that summation by Mathematica Wolfram \(\displaystyle \sum_{x\in A} \sin(\pi x) \) \(\displaystyle A={{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},.........}}\)

how to find that summation by Mathematica Wolfram \(\displaystyle \sum_{x\in A} \sin(\pi x) \) \(\displaystyle A={{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},.........}}\)

undefined MHF Hall of Honor Mar 2010 2,340 821 Chicago May 18, 2010 #2 nice rose said: how to find that summation by Mathematica Wolfram \(\displaystyle \sum_{x\in A} \sin(\pi x) \) \(\displaystyle A={{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},.........}}\) Click to expand... Input: Sum[Sin[Pi/n], {n, 1, Infinity}]. Message: Sum::div: Sum does not converge.

nice rose said: how to find that summation by Mathematica Wolfram \(\displaystyle \sum_{x\in A} \sin(\pi x) \) \(\displaystyle A={{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},.........}}\) Click to expand... Input: Sum[Sin[Pi/n], {n, 1, Infinity}]. Message: Sum::div: Sum does not converge.

C CaptainBlack MHF Hall of Fame Nov 2005 14,972 5,271 someplace May 18, 2010 #3 nice rose said: how to find that summation by Mathematica Wolfram \(\displaystyle \sum_{x\in A} \sin(\pi x) \) \(\displaystyle A={{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},.........}}\) Click to expand... Since for small \(\displaystyle u\) we have \(\displaystyle \sin(u)\approx u\) this is obviously not convergent as the tail of the sum behaves like the harmonic series which is known to be divergent. . CB

nice rose said: how to find that summation by Mathematica Wolfram \(\displaystyle \sum_{x\in A} \sin(\pi x) \) \(\displaystyle A={{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},.........}}\) Click to expand... Since for small \(\displaystyle u\) we have \(\displaystyle \sin(u)\approx u\) this is obviously not convergent as the tail of the sum behaves like the harmonic series which is known to be divergent. . CB