Recent content by gustav1139

Ah. When I said "that guy," I was referring to the first sum in the OP's question. I guess I see how that could be misconstrued.

I apologize. Personification is against the rules?

derp. right.

The first sum, once you've broken them up, is not a geometric sum. Think about what that guy's doing for a little bit.

...##\sum{\frac{1}{n^2}}## converges... doesn't it?

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 Attempt at...

What's ##\sum_{i=1}^{n}3##?

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...

Right you are. All fixed. That (what you said) is however what I'd been working with so that's still no go...

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...