## anonymous one year ago Sigma (summation) notation question. Will post in comment

1. anonymous

$\sum_{k = 0}^{4}\frac{ 1 }{ k^2 + 1 }$ I rewrote it as: $\sum_{k = 0}^{4} (k^2 + 1)^{-1}$ do I use now the formula $\frac{ n(n+1)(2n+1) }{ 6 }$ for the expression (k^2 + 1)^-1 ?

2. anonymous

I think that what I wrote in the previous step is incorrect; I don't think the formula can be applied. It can be solved with brute force. $\frac{ 1 }{ 0^2 + 1} + \frac{ 1 }{ 1^2 + 1} + \frac{ 1 }{ 2^2 + 1} + \frac{ 1 }{ 3^2 + 1} + \frac{ 1 }{ 4^2 + 1}$ $= \frac{ 1 }{ 0 + 1 } + \frac{ 1 }{ 1 + 1 } + \frac{ 1 }{ 4 + 1 } + \frac{ 1 }{ 9 + 1 } + \frac{ 1 }{ 16 + 1 }$ $= \frac{ 1 }{ 1 } + \frac{ 1 }{ 2 } + \frac{ 1 }{ 5 } + \frac{ 1 }{ 10 } + \frac{ 1 }{ 17 } =\frac{ 158 }{ 85 }$

3. anonymous

you are correct in the second solution :)

4. anonymous
5. anonymous

@Halmos I'm I correct that the summation formula won't work? The formulas can be found @ https://mathwiz.uwstout.edu/video/math-156/ch5s1.html 4th video down (It's not necessary to watch the video, it's on the web page).

6. anonymous

the formula u had of k^2 won't work but there is a formula called zeta function and like wise things, this is advanced mathematics don't think about it :P

7. anonymous

Thanks.

8. anonymous

np :)