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tanjung
 3 years ago
simplify ...
tanjung
 3 years ago
simplify ...

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\sum\limits_{n=4}^{25}\frac{1}{n^2+1}\]

tanjung
 3 years ago
Best ResponseYou've already chosen the best response.0i think, the problem wants a rational number

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0hmm http://www.wolframalpha.com/input/?i=sum+%281%2F%7Bn%5E2%2B1%7D%29+from+n%3D4+to+25 well it is rational, , ,

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0if we have interpreted the question correctly, and the wolfram result is the one you are aiming towards, then this wont be a nice problem ,

tanjung
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm... is there a method to calculating series of : 1/(n^2+1) + 1/((n+1)^2+1) + ... + 1/((n+k)^2+1) , @UnkleRhaukus ?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0what is the context that the problem came from ?

tanjung
 3 years ago
Best ResponseYou've already chosen the best response.0this is a question from one of group math also, in my fb. the question ask to calculate or simplify it

wio
 3 years ago
Best ResponseYou've already chosen the best response.0The real question is if there is a simplification of: \[ \sum \frac{1}{n^2+1} \]

wio
 3 years ago
Best ResponseYou've already chosen the best response.0Like, maybe partial fraction decomposition could help us?

wio
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not familiar with this series actually. I'm not sure how to find it's partial sum.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0@calculusfunctions that power series u use is for when \(x<1\)

Stiwan
 3 years ago
Best ResponseYou've already chosen the best response.0@calculusfunctions @mukushla On the topic of the power series, there is this equation: \[\sum_{k=0}^{n}x^k = \frac{1x^{n+1}}{1x}\] If we suppose x < 1 and let n go to infinity, x^(n+1) converges to 0, so the power equation equation becomes \[\sum_{k=0}^{\infty} x^k = \frac{1}{1x}\] 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.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0i meant that power series will not be useful for this problem and there is no way to simplify that expression...if im not wrong

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0Yes @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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