Question

Given that the series as sigma tends to infinity and k=1 of (1/k(k+1)) can be rewritten as sigma tends to infinity and k=1 of (1/k - 1/k+1).. Find the nth partial sum Sn.

Answer 1

\[\sum_{k=1}^{} \frac{ 1 }{ k(k+1) }\]

Answer 2

tends to infinity

Answer 3

Hmmm, I'm wondering if arctan would help with this.

Answer 4

so if k was multiplied in the denom, it would be k^2 +k...I'm going to seek help when I get back to school, Thanks!

Answer 5

Wait are they saying: \[
(1/k(k+1)) = (1/k - 1/k+1)
\]