# Search results for query: *

1. ### Basic Geometric Series Question

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.
2. ### Basic Geometric Series Question

I apologize. Personification is against the rules?

derp. right.
4. ### Basic Geometric Series Question

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.
5. ### Relation of two complex series

...##\sum{\frac{1}{n^2}}## converges... doesn't it?
6. ### Discrete math - simple formalism question

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...
7. ### Relation of two complex series

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...
8. ### Solve Riemann Sum Question for Area Under Curve

What's ##\sum_{i=1}^{n}3##?
9. ### Solve Riemann Sum Question for Area Under Curve

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...
10. ### Solve Riemann Sum Question for Area Under Curve

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})##
11. ### Tricky complex series

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...
12. ### Tricky complex series

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

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.
14. ### Tricky complex series

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...