MHB Find Sum of $\dfrac{4k}{4k^4+1}$

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Sum
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Determine the sum $$\sum_{k=1}^n \dfrac{4k}{4k^4+1}$$.
 
Mathematics news on Phys.org
Here is my solution:

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:

$$S_n=1-\frac{1}{2n^2+2n+1}=\frac{2n(n+1)}{2n^2+2n+1}$$
 
MarkFL said:
Here is my solution:

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)
 
Hey MarkFL, I think I should post the solution which I saw online as well, just to be fair...:o
$$\begin{align*}\sum_{k=1}^n \dfrac{4k}{4k^4+1}&=\sum_{k=1}^n \dfrac{(2k^2+2k+1)-(2k^2-2k+1)}{(2k^2+2k+1)(2k^2-2k+1)}\\&=\sum_{k=1}^n \left(\dfrac{1}{2k^2-2k+1} -\dfrac{1}{2(k+1)^2-2(k+1)+1} \right)\\&=1-\dfrac{1}{2n^2+2n+1}\end{align*}$$

and we're done.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top