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

[anemone]

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

 

[MarkFL]

Here is my solution:

Spoiler

We are given to evaluate:

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

Partial fraction decomposition on the summand allows us to write:

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

Observing that:

\(\displaystyle 2(k+1)^2-2(k+1)+1=2k^2+2k+1\)

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

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

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

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

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

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

 

[anemone]

Here is my solution:

Spoiler

We are given to evaluate:

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

Partial fraction decomposition on the summand allows us to write:

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

Observing that:

\(\displaystyle 2(k+1)^2-2(k+1)+1=2k^2+2k+1\)

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

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

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

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

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

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

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

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

and we're done.

 

[CellCoree]

I would approach this problem by first simplifying the expression to make it easier to work with. We can rewrite the fraction as $\frac{4k}{4k^4+1} = \frac{4k}{(2k^2)^2+1}$. This allows us to see that the denominator is in the form of a perfect square plus 1, which can be factored using the difference of squares formula.

So, we can rewrite the fraction as $\frac{4k}{(2k^2)^2+1} = \frac{4k}{(2k^2+1)(2k^2-1)}$. Now, we can use partial fraction decomposition to write this as $\frac{A}{2k^2+1} + \frac{B}{2k^2-1}$.

Solving for A and B, we get $A = 1$ and $B = -1$. Therefore, we can rewrite the original expression as $\frac{4k}{4k^4+1} = \frac{1}{2k^2+1} - \frac{1}{2k^2-1}$.

Now, we can use the summation property $\sum_{k=1}^n (a_k + b_k) = \sum_{k=1}^n a_k + \sum_{k=1}^n b_k$ to split the summation into two separate summations. This gives us $\sum_{k=1}^n \frac{4k}{4k^4+1} = \sum_{k=1}^n \frac{1}{2k^2+1} - \sum_{k=1}^n \frac{1}{2k^2-1}$.

Now, we can use the summation formula for geometric series $\sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$ to evaluate each summation separately. Plugging in $a = 1$, $r = \frac{1}{2}$, and $n = \infty$, we get $\sum_{k=1}^n \frac{1}{2k^2+1} = \frac{1}{2}\left(1 - \frac{1}{2^n}\right