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Find an expression for

$$\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}.$$

This was given in the chapter for APs. However, I do not see how this relates to them. I tried using telescopic sums, but I am not proficient in them and was thererfore unable to solve this question.

I tried calculating the sum for a few consecutive values, and tried constructing a polynomial that covered them all, but it proved to be difficult.

Any hints on how to solve this question?

Please help.

Community
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DynamoBlaze
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  • I put it into Alpha and got a real mess. Even the sum to infinity is quite messy. Are you sure you have the question correct? – Ross Millikan May 29 '18 at 20:39
  • @RossMillikan My apologies. There seem to be a few very bad typos in the question. – DynamoBlaze May 29 '18 at 20:40
  • @Malay? Do you have any reference of this problem? – imranfat May 29 '18 at 20:41
  • See also : https://math.stackexchange.com/questions/304851/evaluate-sum-limits-k-1-infty-frack2-1k4k21 and https://math.stackexchange.com/questions/2528731/frac1x1-frac2x21-frac4x41-cdots-up-to-n-terms-in and https://math.stackexchange.com/questions/571973/evaluate-sum-limits-n-1-infty-fracnn4n21 and https://math.stackexchange.com/questions/691109/sum-this-series-frac111214-frac212224-ldots-upto-n-terms – lab bhattacharjee May 31 '18 at 07:20

2 Answers2

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Let $a_k = k/(1+k^2+k^4)$. Then:

$$a_k = \underbrace{\frac{k(1-k)}{2(1-k+k^2)}}_{b_k} + \underbrace{\frac{k(1+k)}{2(1+k+k^2)}}_{c_k}.$$

Now observe that

$$b_{k+1} = \frac{(k+1)(-k)}{2(-k+(k+1)^2)} = \frac{-k(1+k)}{2(1+k+k^2)} = -c_k.$$

Can you conclude?

Shaun
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Fimpellizzeri
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  • Beautiful! However, how do I discover that $a_k$ is breakable into two such terms, and then their relation? – DynamoBlaze May 29 '18 at 20:49
  • @Fimpellizieri Please check that I put the bracket in the right place. – Shaun May 29 '18 at 20:50
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    This is called a partial fraction decomposition. The idea to use it... comes with experience, I guess? Since you tagged the question with telescoping series I tried to break the general term in two and see if they satisfied some 'telescoping' relation. – Fimpellizzeri May 29 '18 at 20:50
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    @Shaun You did, thanks! – Fimpellizzeri May 29 '18 at 20:51
  • @Fimpellizieri Partial fractions, eh? Looks like I've got another sleepless night with math for company. Thanks! – DynamoBlaze May 29 '18 at 20:52
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    If you have a polynomial $q(x)$ that can be written as $q(x) = f(x)g(x)$ for lower order polynomials $f$ and $g$, then one can write a rational function $p(x)/q(x)$ as the sum $a(x)/f(x) + b(x)/g(x)$ for some polynomials $a$ and $b$. This is a partial fraction decomposition, and there isn't really much that goes into it besides usual polynomial long division. – Fimpellizzeri May 29 '18 at 20:55
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A telescoping series:

$\sum_{n=1}^{k} f(n) - f(n+1) = f(1) - f(k+1)$

How do we get our series into that form?

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

Here is the really tricky bit:

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

$\sum_{n=1}^{k}\frac {n}{n^4 + n^2 + 1} = \frac 12 - \frac {1}{2(k^2+k+1)}$

Doug M
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  • How did you get the third step? Could you please elaborate? – DynamoBlaze May 29 '18 at 21:00
  • The most direct way is to say $\frac {n}{(n^4+n^2+1)} = \frac {An+B}{n^2 - n + 1} + \frac {Cn+D}{n^2+n + 1} \implies n = (An+B)(n^2+n+1) + (Cn+D)(n^2-n+1)$ and solve for $A,B,C, D.$ But if you start with the guess that $(An+B) = -(Cn+D)$ you see alot of what cancels (and what doesn't) right away, but that a solution exists, saving some algebra. – Doug M May 29 '18 at 21:05