Series Convergence

By Yatir Halevi on Friday, January 25, 2002 - 11:05 am:

If I have two series:
(1) 1/x₁ +1/x₂ +...
(2) 1/y₁ +1/y₂ +...

They both converge.

Does the series:
(3) 1/(y₁ x₁ )+1/(y₂ x₂ )+...
converge?

If I know that (1) converges to L and that (2) Converges two P, To what (3) converges (if it does) and is it expressable by P and L?

Thanks,
Yatir

By Michael Doré on Friday, January 25, 2002 - 12:56 pm :

This depends on whether you restrict $x_{n}$ , $y_{n}$ to being positive. If you do make this restriction then since $\sum 1 / x_{i}$ converges then $1 / x_{i} \to 0$ as $i \to \infty$ . In particular, there exists $n$ such that for all $i > n$ we have $1 / x_{i} < 1$ . This means that for all $i > n$ we have $1 / (y_{i} x_{i}) < 1 / y_{i}$ and so $1 / (y_{n + 1} x_{n + 1}) + 1 / (y_{n + 2} x_{n + 2}) + \dots$ converges so $\sum 1 / (y_{i} x_{i})$ converges.

If you don't restrict $x_{i}$ , $y_{i}$ to being positive and only restrict them to being non-zero, then the result is false.

For example set $x_{i} = y_{i} = (- 1)^{i} \sqrt{i}$ then $\sum 1 / x_{i}$ and $\sum 1 / y_{i}$ converge (by the alternating series test) yet $\sum 1 / (x_{i} y_{i})$ diverges.

Finally, even if $\sum 1 / (x_{i} y_{i})$ converges, you cannot express this in terms of $P$ and $L$ . For example take $x_{i} = y_{i} = 1 / 2^{i}$ then $P = L = 1$ and the sum in (3) is 1/3. However if you take $x_{i} = y_{i} = 3 / (4^{i})$ then $P = L = 1$ still yet the sum in (3) is 51/17 or something, which is not 1/3. In other words even if the values of $P$ and $L$ are kept the same, it is possible for the sum in (3) to change.

By Yatir Halevi on Friday, January 25, 2002 - 05:14 pm:

Thanks Michael

Yatir