The summation in question:
$$\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}$$
I have been able to factorize $i^4+i^2+1$ as $(i^2+i+1)(i^2-i+1)$ but I doubt this will help.
What is the solution?
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2are you meant to come up with a formula? or why can you not simply add up the 10 terms? – Chinny84 Nov 25 '16 at 13:54
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@Chinny84 I need a proper solution using rules of summation. Using brute force isn't allowed – Mihir Chaturvedi Nov 25 '16 at 13:56
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You might like the discussion on this post: http://math.stackexchange.com/questions/2029003/a-nice-infinite-series-sum-limits-n-1-infty-frac1nn4n21-frac re: ${\sum\limits_{n=1}^\infty}\frac{1}{n!(n^4+n^2+1)}$ – Mehness Nov 25 '16 at 14:05
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Writting $$\sum^{10}_{i=1}\frac{i}{i^4+i^2+1} = \frac{1}{2}\sum^{10}_{i=1}\bigg[\frac{(i^2+i+1)-(i^2-i+1)}{(i^2+i+1)(i^2-i+1)}\bigg]$$
$$ = \frac{1}{2}\sum^{10}_{i=1}\bigg[\frac{1}{i^2-i+1}-\frac{1}{i^2+i+1}\bigg]$$
Now Using Telescopic Sum (expanding summation), Then we get
$$ =\frac{1}{2}\bigg[\frac{1}{1^2-1+1}-\frac{1}{10^2+10+1}\bigg]= \frac{1}{2}\left(1-\frac{1}{111}\right) = \frac{55}{111}$$
juantheron
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1It might help the OP if you were more explicit about why all the middle terms cancel out. – Ian Miller Nov 25 '16 at 14:04
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@juantheron I don't think that the middle terms can be cancelled out this way... – Mc Cheng Nov 25 '16 at 14:06
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1I meant write $\frac{1}{i^2-i+1}-\frac{1}{i^2+i+1}$ as $\frac{1}{i(i-1)+1}-\frac{1}{(i+1)i+1}$ so it is clearly in the form of $f(i)-f(i+1)$. – Ian Miller Nov 25 '16 at 14:07
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