- 10 Apr '08 00:02

1+1+1+1+1+...*Originally posted by adam warlock***On the convergence of series. So you have an infinite summation and you know that the difference between sucessive terms goes to 0 ans n goes to infinity. Does this mean that the series converge? If so prove it if not give a counter-example.**

The difference between sucessive terms goes to 0 as n goes to infinity and the series diverges. - 10 Apr '08 09:43 / 1 edit

as already pointed out, giving the example of the harmonic series, that is 1 + 1/2 + 1/3 + 1/4 etc. your condition is*Originally posted by adam warlock***On the convergence of series. So you have an infinite summation and you know that the difference between sucessive terms goes to 0 ans n goes to infinity. Does this mean that the series converge? If so prove it if not give a counter-example.****necessary**but not**sufficient**. In lamest terms, the difference between two consecutive terms has to not only go to zero, it has to go to zero "fast" enough

The proof that the harmonic series doesn't converge is actually quite simple

think of 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 ... now group the terms as follows 1 + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ( 1/9 + ... + 1/16 ) + ( 1/17 + ... + 1/32 ) + ...

then you can see that if you sum up the terms grouped together, you will always get a number at least as big as 1/2. So the partial sums of the harmonic series are greater than the partial sums of 1 + 1/2 + 1/2 + 1/2 ... which clearly diverges, thus harmonic series also diverges.

Yet the different between two consecutive terms 1/n, 1/n+1 is 1/n(n+1)

and the sequence 1/n(n+1) clearly converges to zero. - 10 Apr '08 10:00

something like this an = 1^n or an = 1 + 1/n, etc.*Originally posted by David113***What do you mean "general terms with n's on it"?**

so clearly the condition that the difference between the terms goes to zero is very loose. You could improve it by saying that the**sequence**of the terms converges to zero. That condition would be better, but still not sufficient (as pointed out above).

The sufficient condition is that the sequence of**partial sums**of the series has to converge to 0.

A term Sn in the sequence S of partial sums of a series a0 + a1 + a2 ...

is given as Sn = sum{0, n} an, so S1 = a0, S2 = a0 + a1, etc ... Moverover, the limit of the sequence is equal to the sum of the series. - 10 Apr '08 10:25

I meant general*Originally posted by David113***What do you mean "general terms with n's on it"?****term**with an n in it. Sorry for the extra s. But I think I might be using a term that isn't used in english. As I said in the other thread I was thought that whas is being summed in a series can be translated as "general term" to english but maybe that isn't the right technical term. But I was looking for something that was already posted. And when I was asked this question that was the answer I gave. But I liked your lateral thinking. You gave an example of a pretty straightforward counter-example.

Ps: Went to google and searched for general term and the first hit I got was this: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2005;task=show_msg;msg=1941 So it does seem that my free translantion was right. - 10 Apr '08 10:26

Either that or the Cauchy condition. u_{n+p}-u_n goes to 0 as n goes to infinity for any value of p.*Originally posted by 3v1l5w1n***something like this an = 1^n or an = 1 + 1/n, etc.**of the terms converges to zero. That condition would be better, but still not sufficient (as pointed out above).

so clearly the condition that the difference between the terms goes to zero is very loose. You could improve it by saying that the [b]sequence

The sufficient condition is that the s ...[text shortened]... 0, S2 = a0 + a1, etc ... Moverover, the limit of the sequence is equal to the sum of the series.[/b]