I was doing something and I got $\displaystyle \sum_{n=1}^{m}arctan\bigg(\frac{1}{n^2+n+1}\bigg)= arctan(m+1)-\frac{\pi}{4}$...does anyone concur?
Last edited by Mathstud28; Apr 12th 2008 at 07:17 PM.
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Originally Posted by Mathstud28 I was doing something and I got $\displaystyle \sum_{n=1}^{m}arctan\bigg(\frac{1}{n^2+n+1}\bigg)= arctan(m+1)-\frac{\pi}{5}$...does anyone concur? It's not true for m = 1 or m = 2 .... Did you do such a check?
care to post your method or the complete question ?
Originally Posted by Mathstud28 I was doing something and I got $\displaystyle \sum_{n=1}^{m}arctan\bigg(\frac{1}{n^2+n+1}\bigg)= arctan(m+1)-\frac{\pi}{4}$...does anyone concur? Now that I fixed my typo is this correct?
Originally Posted by Mathstud28 Now that I fixed my typo is this correct? An improvement.
Originally Posted by mr fantastic An improvement. Does that mean I am correct?
Originally Posted by Mathstud28 Does that mean I am correct? Well, it means it's correct for m = 1 and m = 2 ..... I haven't done a detailed analysis - perhaps you could post all your working for possible critique.
Originally Posted by mr fantastic Well, it means it's correct for m = 1 and m = 2 ..... I haven't done a detailed analysis - perhaps you could post all your working for possible critique. I think I will tomorrow its a lot of work..I think I am right =D....at least up to m=10
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