# Find expression for sum of series

I want to find a formula for the sum of this series using its general term. How to do it?

Series $$S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}$$

General Term

$$S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}$$

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I changed n\ \mbox{terms} to n\text{ terms}. The mismatch in sizes of fonts resulted from use of an incorrect method. –  Michael Hardy Jul 12 '14 at 22:23

Using partial fractions, we get $\dfrac{i}{i^4+i^2+1} = \dfrac{\tfrac{1}{2}}{i^2-i+1} - \dfrac{\tfrac{1}{2}}{i^2+i+1}$.

Thus, $S_n = \displaystyle\sum_{i = 1}^{n}\dfrac{i}{i^4+i^2+1} = \dfrac{1}{2}\sum_{i = 1}^{n}\dfrac{1}{i^2-i+1} - \dfrac{1}{i^2+i+1}$.

Since $(i+1)^2-(i+1)+1 = i^2+i+1$, this sum telescopes to $S_n = \dfrac{1}{2}\left(1 - \dfrac{1}{n^2+n+1}\right)$.

Taking the limit as $n \to \infty$ gives $S = \dfrac{1}{2}$.

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thank you so much,but i have one doubt ,in which cases can i apply partial fraction to calculate sum? –  user3481652 Jul 12 '14 at 20:47
Anytime you have a rational function (whose numerator has a smaller degree than its denominator), you can always try using partial fractions. Whether or not it will lead to a solution depends on the particular problem. –  JimmyK4542 Jul 12 '14 at 20:49
okay thank you for your help :) –  user3481652 Jul 12 '14 at 20:51
Note that the factorisation $i^4+i^2+1=(i^2+i+1)(i^2-i+1)$ comes easily if you notice that $(i^2-1)(i^4+i^2+1)=i^6-1=(i^3+1)(i^3-1)=(i+1)(i^2-i+1)(i-1)(i^2+i+1)$ –  Mark Bennet Jul 12 '14 at 21:06

Use partial fractions,

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

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Write it out:

$$\begin{eqnarray} \sum_{i=1}^\infty \frac{i}{i^4 + i^2 + 1} &=& \sum_{i=1}^\infty \left( \frac{2}{4i^2 - 4i + 4} - \frac{2}{4i^2 + 4i + 4}\right)\\ &=& \sum_{i=1}^\infty \left( \frac{2}{\Big(2i-1\Big)^2 + 3} - \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\ &=& \frac{2}{\Big( 2 - 1\Big)^2 + 3} + \sum_{i=1}^\infty \left( \frac{2}{\Big(2i+1\Big)^2 + 3} - \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\ &=& \frac{1}{2}. \end{eqnarray}$$

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Mind to give reason for downvote? –  johannesvalks Jul 12 '14 at 21:06
I didn't downvote you. But I think you posted the solution AFTER other users. I think they care about "speed" + "accuracy"....next time... –  NotALoner Jul 12 '14 at 21:35
Well - that is the problem when you reply a post and in the mean time you watch the worldcup so you are late with the final result.... Thanks for the tip / advice!! –  johannesvalks Jul 12 '14 at 21:39
@ user2584283 I think the OP asks for (a formula for) the sum of the series. –  user84413 Jul 13 '14 at 1:01
If you look at the question's edit history, you will see that at one time, it asked for the sum of the infinite series, not just the first $n$ terms. –  JimmyK4542 Jul 13 '14 at 1:41