7

Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$

MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtraction was written in numerator. so I did the same thing as we do in the method of partial fraction, and started putting $1,2,3$ and so on. my answer came didn't match with the right answer.

AccidentalFourierTransform
  • 2,592
Priyanka Vithani
  • 431
  • Can you show the exact steps you wrote down? It would be a great help to make our own judgements from the steps you've written down, rather than you analysing them for us. – Toby Mak Jul 13 '17 at 12:37

5 Answers5

6

We can write the sum as:

$$\sum _{r=1}^{\infty} \frac{8r}{4r^4+1}$$

$$= \sum _{r=1}^{\infty} \frac{8r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$$

$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2r^2 + 2r + 1}$$

$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2(r + 1)^2 - 2(r + 1) + 1}$$

$$=\frac{2}{2\cdot1^2 - 2\cdot1 + 1}$$

$$ = 2$$

Josh
  • 945
4

Hint

$$4(r)^4+1=(2r^2+1)^2-(2r)^2=(2r^2-2r+1)(2r^2+2r+1)$$

and use fraction partial decomposition.

Arnaldo
  • 21,342
2

Hint. The sum is telescopic. Note that $4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$, and $$\frac{8r}{4r^4+1}=\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}.$$ Hence $$\sum_{r=1}^n\frac{8r}{4r^4+1}=\sum_{r=1}^n\left(\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}\right)=\frac{2}{2\cdot 1(1-1)+1}-\frac{2}{2(n+1)n+1}.$$ Can you find the sum of the series now?

Robert Z
  • 145,942
2

Let the general term be $\frac {8r}{4r^4+1}=\frac {2r}{r^4+(\frac {1}{2})^2}=\frac {2r}{(r^2+\frac {1}{2})^2-r^2} =\frac {2r}{(r^2+\frac {1}{2}-r)(r^2+\frac {1}{2}+r)} $ now $2r=r^2+\frac {1}{2}+r-(r^2+\frac {1}{2}-r) $ so iur series is a telescoping one and is equal to $S=\sum _0^{\infty} \frac {1}{r^2+\frac {1}{2}-r}-\frac {1}{r^2+\frac {1}{2}+r}=2$

Archis Welankar
  • 15,910
2

We can use partial fraction decomposition to write this as \begin{align}\sum_{n=1}^\infty \frac{8n}{4n^4+1}&=\sum_{n=1}^\infty\left(\frac{2}{2 n^2 - 2 n + 1} - \frac{2}{2 n^2 + 2 n + 1}\right)\end{align}

If we start writing out terms of this sequence, we get \begin{array}{ccccccccccc}(n=1)&2&-&\frac{2}{5}\\ (n=2)&&+&\frac 25&-&\frac 2{13}\\ (n=3)&&&&+&\frac 2{13}&-&\frac 2{25}\\ &&&&&&+&\cdots\\ (n=\infty)&&&&&&&+&\frac{2}{2\infty^2-2\infty+1}&-&\frac{2}{2\infty^2+2\infty+1}\end{array}

We therefore have a telescoping sequence, so the sum is:

$$\sum_{n=0}^\infty \frac{8n}{4n^4+1}=2-\frac{2}{2\infty^2+2\infty+1}$$

We can see that, $$\lim_{n\to\infty}\dfrac{2}{2n^2+2n+1}=0$$

Therefore, $$\sum_{n=1}^\infty\frac{8n}{4n^4+1} = 2$$

lioness99a
  • 4,943