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Infinite Series: sigma n^2/(n^2 +1) 
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#1
Apr2407, 11:29 PM

P: 35

If I take the limit on the sum... I get 1/1 = 1
If the limit does NOT = 0 then sigma f(x) diverges...I'm not quite sure I follow this... Does this mean that in order for the equation to converge, the sum (sigma) must be = to 0? 


#2
Apr2407, 11:49 PM

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P: 4,771

You're again confusing the limit of the argument (here n^2/(n^2 +1)) with the actual sum, which is the limit of
[tex]\sum_{n=1}^N\frac{n^2}{n^2+1}[/tex] as N>infty. The theorem is saying that if the limit of the argument is not 0, then you must conclude that the sum diverges. If it IS 0, then you cannot conclude anything: the sum could converge or diverge. For instance, consider the old harmonic series [tex]\sum\frac{1}{n}[/tex] Sure, 1/n>0 but it is well known that the harmonic series diverges nonetheless. 


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