Finding $\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}$.

Question

Find an expression for

$$\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}.$$

This was given in the chapter for APs. However, I do not see how this relates to them. I tried using telescopic sums, but I am not proficient in them and was thererfore unable to solve this question.

I tried calculating the sum for a few consecutive values, and tried constructing a polynomial that covered them all, but it proved to be difficult.

Any hints on how to solve this question?

Please help.

Answer 1

Let $a_k = k/(1+k^2+k^4)$. Then:

$$a_k = \underbrace{\frac{k(1-k)}{2(1-k+k^2)}}_{b_k} + \underbrace{\frac{k(1+k)}{2(1+k+k^2)}}_{c_k}.$$

Now observe that

$$b_{k+1} = \frac{(k+1)(-k)}{2(-k+(k+1)^2)} = \frac{-k(1+k)}{2(1+k+k^2)} = -c_k.$$

Can you conclude?

Answer 2

A telescoping series:

$\sum_{n=1}^{k} f(n) - f(n+1) = f(1) - f(k+1)$

How do we get our series into that form?

$(n^4 + n^2 + 1) = (n^2+n +1)(n^2-n+1)\\ \frac {n}{(n^4 + n^2 + 1)} = \frac {1}{2(n^2 - n + 1)} - \frac {1}{2(n^2 + n + 1)}$

Here is the really tricky bit:

$\frac {1}{(n^2 + n + 1)} = \frac {1}{(n+1)^2 - (n+1) + 1}$

$\sum_{n=1}^{k}\frac {n}{n^4 + n^2 + 1} = \frac 12 - \frac {1}{2(k^2+k+1)}$