Question

Sigma (summation) notation question.

Will post in comment

Answer 1

\[\sum_{k = 0}^{4}\frac{ 1 }{ k^2 + 1 }\]

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 ?

Answer 2

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.

\[\frac{ 1 }{ 0^2 + 1} + \frac{ 1 }{ 1^2 + 1} + \frac{ 1 }{ 2^2 + 1} + \frac{ 1 }{ 3^2 + 1} + \frac{ 1 }{ 4^2 + 1}\]
\[= \frac{ 1 }{ 0 + 1 } + \frac{ 1 }{ 1 + 1 } + \frac{ 1 }{ 4 + 1 } + \frac{ 1 }{ 9 + 1 } + \frac{ 1 }{ 16 + 1 }\]
\[= \frac{ 1 }{ 1 } + \frac{ 1 }{ 2 } + \frac{ 1 }{ 5 } + \frac{ 1 }{ 10 } + \frac{ 1 }{ 17 } =\frac{ 158 }{ 85 } \]

Answer 3

you are correct in the second solution :)

Answer 4

http://www.wolframalpha.com/input/?i=%5Csum_%7Bk+%3D+0%7D%5E%7B4%7D%5Cfrac%7B+1+%7D%7B+k%5E2+%2B+1+%7D+

Answer 5

@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).

Answer 6

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

Answer 7

Thanks.

Answer 8

np :)