The 'implies' is there to emphasize the logical connection there. If you have two reals like that, then you can find such a z. Moreover, if you can't find such a z, then x=y (or one of x,y is not a real number, which seems less likely).
I probably would have left out the arrow as well, since the...
Homework Statement
Suppose that ##\left\{a_n\right\}## is a sequence of complex numbers with the property that ##\sum{a_n b_n}## converges for every complex sequence ##\left\{b_n\right\}## such that ##\sum{|b_n|^2}<\infty##. Show that ##\sum{|a_n|^2}<\infty##.
Homework Equations
The...
There's two mistakes I see:
when you plugged in f(2+12/n), you flipped a sign.
you can't really "pull out 36/n" the way you have; sorry I didn't catch this first time around.
When you're at the stage where you have ##\frac{12}{n}\sum{(3-6i/n)}##, you have to be a bit more careful about how...
Yeah you have the formula for the Riemann sum just slightly (but crucially) wrong. It should be:
##\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(a+\frac{i(b-a)}{n})\cdot(\frac{b-a}{n})##
Thanks! I think that broke it open. I had been restricting myself unnecessarily with what I could assume if ##\sum|a_i|## diverged.
I picked ##b_k = \frac{\overline{a_i}}{c_i |a_i|}## where the c's are the largest n such that i > k_n. Then the b's go to 0 since they do in modulus, and you...
Thanks! I'll give that a try, but isn't it an issue that they're complex sequences, if I want to use the MVT?
Also, why can you assume a(x) is a nonnegative real? Or perhaps I'm misunderstanding what you mean by resolve.
Homework Statement
Suppose that \left\{a_{n}\right\} is a sequence of complex numbers with the property that \sum{a_{n}b_{n}} converges for
every complex sequence \left\{b_{n}\right\} such that \lim{b_{n}}=0. Prove that \sum{|a_{n}|}<\infty.
Homework Equations
The Attempt at a Solution
I...