Does a diverging series always have to diverge to ± infinity?

Question

In my Calculus textbook it doesn't say.

I'm asking because of a specific question it gave, where the series x-(x^2)/2+(x^3)/3+....(-1)^n(x)^(n+1)/(n+1) "appears" to converge to ln2. Only I don't think it does. Is it possible that they used the word "appears" because it is only that, it just appears to converge. Or is it impossible for a series to diverge around a certain number?

Answer 1

All alternating series (+-+-+ terms) converge if the terms are getting smaller. However, this is not the case for your series. Your terms are geting bigger provided x>1 but it may still converge. In this case it does as your series converges to ln(x+1). So, it appears that if x=1 then the series converges to ln(2). However, it holds only for -1

Answer 2

Diverging Series

Answer 3

When the word "diverging" is used, it is implied that the series is "converging to plus-minus infinity". The choice of words depends on your setup of topology. I have had some books (and professors) refer to it as diverging to infinity. Some say converging to infinity (treating infinity like a number). Some just say diverging.

BUT, when the word "diverging" is used, it means that it is increasing (decreasing) without bound which means diverging to infinity.

As for your example, this is PRECISELY the reason everything is proved in math. In sciences (including math) some of the results are "obvious" and intuitive. Others however are competely unintuitive and that is why we must prove if the series converges or not and if yes to what value.

I mean here is another example. The series 1/n seems like it would converge (because the individual terms) are strictly decreasing BUT the series does diverge.

There are other possibilities however. For example, the sequence sin(n). It is NOT divergin because it is bounded from above (+1) and below (-1) but as n goes to infinity, this sequence doesn't converge either because it is constantly oscillating.

Convergence and divergence are not the only possibilities. For the case of sin(n) above, we just say that the limit doesn't exist.

Answer 4

You are getting a lot of wrong answers here. it is quite possible for a series to diverge but have the partial sums not go to infinity. just consider
sum (-1)^n.

This series has partial sums are 1,0,1,0,1,0...
which is a sequence that does not converge, so the series diverges. However, the partial sums do not go to infinity. Sorry Wilhelm, you are wrong here. The sum of sin(n) *diverges*.

In your example, when x=1, the series *does* converge and, in fact, it converges to ln(2). However, to *prove* this is not so easy. It does not follow from the usual versions of Taylor's theorem.

Answer 5

Yes, a divergent series is defined as one that doesn't approach a finite number but goes on increasing in magnitude and approaching infinity.

Hence, all diverging series always diverge to + or -infinity