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 $\frac{1}{2}$
The formula for this sum is ∑n/[n^4+n^2+1] = ∞
To evaluate this sum, we use the formula ∑n/[n^4+n^2+1] = ∞.
No, this sum cannot be simplified further as it is already in its simplest form.
The limit of this sum as n approaches infinity is infinity (∞).
This sum is often used in mathematical analysis to illustrate the concept of infinite series and the behavior of sums as n approaches infinity.