Question regarding infinite series.?

Question

Let "sum" be the sum of a series from n=1 to infinite.

I understand that sum(a-b) = sum(a) - sum (b)

If you are testing for convergence/divergence and find that sum(a) is divergent, does that imply that sum(a-b) is also divergent. Or do the way a and b interact affect divergence/convergence.

Thanks for the help.

Answer 1

Without knowing the function, you can't generalize about it.

If ƒ(a) or ƒ(b) diverge, ƒ(a±b) might converge or might not.


Here are some examples, using ƒ(x) = sum(xⁿ):
|--: ƒ(x) converges for |x| < 1
|--: ƒ(x) diverges for 1 ≤ |x|

Take a = 3.14159, b=2.71828:
ƒ(a) diverges; ƒ(b) diverges; ƒ(a-b) converges.

Take a = 1.4142, b=.5:
ƒ(a) diverges; ƒ(b) converges; ƒ(a-b) converges.

Take a = 0.61803, b=1.33333:
ƒ(a) converges; ƒ(b) diverges; ƒ(a-b) converges.

Take a = -0.693147, b=.86603:
ƒ(a) converges; ƒ(b) converges; ƒ(a-b) diverges.

Answer 2

sometimes sum(a-b) can be shown to be convergent even if both sum(a) and sum(b) are divergent. The easy example in the first answer shows this

This is a hard question to answer properly since it's not in the usual notation. Then again I don't have the fonts to do that, either, or I could come up with some good examples

Answer 3

Well, the answer to the first part is obviously no. Like if a=b and a is diverfent, a-b=0 and is therefore convergent.

So normally, you can't say much. What is sure is that if both, a and b are convergent, a+b and a-b will also be so. Also if one and only one is divergent (the other being convergent) the a+b and a-b will be divergent. Hopefully, I'm not mistaken, but it seems true.
Good luck!

Answer 4

The sum of an limitless series is merely defined if it converges on a unmarried value. This alternates between a million and 0, so it would not converge - the sum of the series is undefined.