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Topic:
Can two series, both diverges, multiplied give a series that converges?
Replies:
22
Last Post:
Oct 7, 2017 12:52 AM




Re: Can two series, both diverges, multiplied give a series that converges?
Posted:
Oct 6, 2017 1:00 PM


fredag 6. oktober 2017 18.05.49 UTC+2 skrev Mike Terry følgende: > On 06/10/2017 16:21, konyberg wrote: > > fredag 6. oktober 2017 17.13.31 UTC+2 skrev Peter Percival følgende: > >> konyberg wrote: > >>> Consider these two series. s = lim (n=1 to inf) Sum(1/n) and t = lim > >>> (n=1 to inf) Sum(1/(1+n)). Both series diverges, going to infinity. > >>> Now if we multiply these, > >> > >> What is the definition of the product of two infinite series? > >> > >> > >>> we can argue that every product of the new > >>> series is smaller or equal to 1/n^2. So it should converge. Or can > >>> we? Let us write the first as a series without the sigma and the > >>> other with sigma. s*t = (1+1/2+1/3+ ...) * t. And since the first > >>> from s (1 * t) diverges, how can s*t converge? > >>> > >>> KON > >>> > > It is the multiplication of the two series. > > That doesn't answer Peter's question. Each series has infinitely many > terms, and you need to say what you mean the product to be calculated > from those terms. > > If you thought this through carefully, you'd realise straight away the > answer to your original question, I think! > > To get you started in the right direction, suppose the first series is: > > Sum [n=1 to oo] (a_n) > > and the second is: > > Sum [n=1 to oo] (b_n) > > Now, what do you mean by the "product" of these series? > > If you feel tempted to reply "just multiply them together", then ask > yourself "multiply WHAT together exactly?" (Remember, multiplication is > an operation that takes TWO numbers, and gives a single number as the > answer. In the two series, you have INFINITELY many numbers...) > > Or perhaps your answer will be that the product of the two series is > some new third series? (If so, then say what is the n'th term of this > new series?) > > > Regards, > Mike.
There are two ways to give the product: 1: sum(a) * sum (b) = sum (c) or 2: every of sum a multiplied with the ones of sum of b. If the sum of any of them is not (in 2) as we like, then none of them are! KON



