Infinite Series: sigma n^2/(n^2 +1)

[anderma8]

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?

 

[quasar987]

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.

 

[tacman]

No, it does not mean that the sum must be equal to 0 in order for the series to converge. The statement is saying that if the limit of the terms in the series (in this case, n^2/(n^2+1)) does not approach 0 as n approaches infinity, then the series will diverge. In other words, if the terms in the series do not get smaller and smaller as n increases, then the series will not converge. In this case, since the limit of n^2/(n^2+1) is equal to 1, the series will diverge. This means that the sum of the infinite number of terms in the series will not have a finite value.