The thread discusses the evaluation of the infinite sum \(\sum_{n = 0}^{\infty}\frac{n}{n^4+n^2+1}\). The focus is on the mathematical reasoning and techniques that could be applied to solve this sum.
Participants generally agree on the interest in evaluating the sum, but no specific solutions or methods have been proposed, leaving the discussion open-ended.
No mathematical steps or assumptions have been detailed, and the evaluation remains unresolved.
Readers interested in series summation, mathematical analysis, or those looking for collaborative problem-solving in mathematics may find this discussion relevant.
lfdahl said:Evaluate the sum:\[\sum_{n = 0}^{\infty}\frac{n}{n^4+n^2+1}\]
kaliprasad said:as 1st term is zero so the um can be taken from 1 to infinite
the n$th$ term is $\frac{n}{n^4+n^2+1}= \frac{n}{n^4+2n^2+1-n^2}= \frac{n}{(n^2+1)^2-n^2}= \frac{n}{(n^2+n+1) (n^2 -n + 1)} $
$=\frac{1}{2}\frac{2n}{(n^2+n+1)(n^2 -n + 1)} $
$=\frac{1}{2}\frac{(n^2+n+1)- (n^2 -n + 1)}{(n^2+n+1)(n^2 -n + 1)} $
$=\frac{1}{2}(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}) $
$=\frac{1}{2}(\frac{1}{n(n-1)+1}-\frac{1}{n(n+1)+1}) $
so summing from 1 to infinite we get
sum = $\sum_{1}^\infty(\frac{1}{2}(\frac{1}{n(n-1)+1}-\frac{1}{(n+1)n+1})$
$=\frac{1}{2}(\sum_{1}^\infty(\frac{1}{n(n-1)+1}-\frac{1}{(n+1)n+1})$
$=\frac{1}{2}(\sum_{1}^\infty(\frac{1}{n(n-1)+1})-\sum_{1}^\infty(\frac{1}{(n+1)n+1})$
$=\frac{1}{2}(\sum_{1}^\infty(\frac{1}{n(n-1)+1})-\sum_{2}^\infty(\frac{1}{(n-1)n+1})$
now all the terms except 1st term cancell leaving $=\frac{1}{2}(\frac{1}{1(1-1)+1})$ or $\frac{1}{2}$