simplify ...

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|dw:1356426308338:dw|
\[=\sum\limits_{n=4}^{25}\frac{1}{n^2+1}\]
i think, the problem wants a rational number

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hmm http://www.wolframalpha.com/input/?i=sum+%281%2F%7Bn%5E2%2B1%7D%29+from+n%3D4+to+25 well it is rational, , ,
if without wolfram ?
if we have interpreted the question correctly, and the wolfram result is the one you are aiming towards, then this wont be a nice problem ,
hmmm... is there a method to calculating series of : 1/(n^2+1) + 1/((n+1)^2+1) + ... + 1/((n+k)^2+1) , @UnkleRhaukus ?
im not sure ,
what is the context that the problem came from ?
this is a question from one of group math also, in my fb. the question ask to calculate or simplify it
The real question is if there is a simplification of: \[ \sum \frac{1}{n^2+1} \]
Like, maybe partial fraction decomposition could help us?
I'm not familiar with this series actually. I'm not sure how to find it's partial sum.
@calculusfunctions that power series u use is for when \(|x|<1\)
@calculusfunctions @mukushla On the topic of the power series, there is this equation: \[\sum_{k=0}^{n}x^k = \frac{1-x^{n+1}}{1-x}\] If we suppose |x| < 1 and let n go to infinity, x^(n+1) converges to 0, so the power equation equation becomes \[\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}\] So you can use the power series only to calculate infinite series if |x|<1, but as the sum is finite in this example we could use it as well, if we would be able find a constant factor x so that \[\frac{1}{n²+1} * x =\frac{1}{(n+1)²+1} \forall n \in [4,24] \subseteq \mathbb{N}\] You see immediately that this factor x exists but it is not CONSTANT (it depends on n), so we can't use the power series here. @tanjung I'm pretty sure that the answer UncleRauhkus gave you is the correct one, because I can't see the point of an exercise that would require you to do any calculations beside that, because it certainly would not "simplify" the matter.
i meant that power series will not be useful for this problem and there is no way to simplify that expression...if im not wrong
Yes @mukushla and @sitwan you're both right. I know that, that's why I didn't proceed further. I was just thinking out loud.

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