Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
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 at 22:23

3 Answers 3

up vote 8 down vote accepted

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 $S = \dfrac{1}{2}$.

share|improve this answer
thank you so much,but i have one doubt ,in which cases can i apply partial fraction to calculate sum? –  user3481652 Jul 12 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 at 20:49
okay thank you for your help :) –  user3481652 Jul 12 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 at 21:06

Use partial fractions,

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

share|improve this answer

Write it out:

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

share|improve this answer
Mind to give reason for downvote? –  johannesvalks Jul 12 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... –  ah-huh-moment. Jul 12 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 at 21:39
@ user2584283 I think the OP asks for (a formula for) the sum of the series. –  user84413 Jul 13 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 at 1:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.