MHB Find Sum of $\dfrac{4k}{4k^4+1}$

[anemone]

Determine the sum $$\sum_{k=1}^n \dfrac{4k}{4k^4+1}$$.

 

[MarkFL]

Here is my solution:

Spoiler

We are given to evaluate:

$$S_n=\sum_{k=1}^n\left(\frac{4k}{4k^4+1} \right)$$

Partial fraction decomposition on the summand allows us to write:

$$S_n=\sum_{k=1}^n\left(\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} \right)$$

Observing that:

$$2(k+1)^2-2(k+1)+1=2k^2+2k+1$$

and using the rule of linearity of the summand and re-indexing the first sum, we obtain:

$$S_n=\sum_{k=0}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^n\left(\frac{1}{2k^2+2k+1} \right)$$

Pulling the first term from the first sum and the last term from the second sum, we may write:

$$S_n=1+\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\frac{1}{2n^2+2n+1}$$

The two sums add to zero, and we are left with:

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

 

[anemone]

Here is my solution:

Spoiler

We are given to evaluate:

$$S_n=\sum_{k=1}^n\left(\frac{4k}{4k^4+1} \right)$$

Partial fraction decomposition on the summand allows us to write:

$$S_n=\sum_{k=1}^n\left(\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} \right)$$

Observing that:

$$2(k+1)^2-2(k+1)+1=2k^2+2k+1$$

and using the rule of linearity of the summand and re-indexing the first sum, we obtain:

$$S_n=\sum_{k=0}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^n\left(\frac{1}{2k^2+2k+1} \right)$$

Pulling the first term from the first sum and the last term from the second sum, we may write:

$$S_n=1+\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\frac{1}{2n^2+2n+1}$$

The two sums add to zero, and we are left with:

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

Awesome, MarkFL! (Nerd)And thanks for participating!(Sun)

 

[anemone]

Hey MarkFL, I think I should post the solution which I saw online as well, just to be fair...

Spoiler

$$\begin{align*}\sum_{k=1}^n \dfrac{4k}{4k^4+1}&=\sum_{k=1}^n \dfrac{(2k^2+2k+1)-(2k^2-2k+1)}{(2k^2+2k+1)(2k^2-2k+1)}\\&=\sum_{k=1}^n \left(\dfrac{1}{2k^2-2k+1} -\dfrac{1}{2(k+1)^2-2(k+1)+1} \right)\\&=1-\dfrac{1}{2n^2+2n+1}\end{align*}$$

and we're done.