# Thread: Summation

1. ## Summation

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?

2. 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?

3. care to post your method or the complete question ?

4. ## Ok

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?

5. Originally Posted by Mathstud28
Now that I fixed my typo is this correct?
An improvement.

6. ## I am sorry

Originally Posted by mr fantastic
An improvement.
Does that mean I am correct?

7. 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.

8. ## Ehh

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