# Math Help - Complex Analysis, Sequence

1. ## Complex Analysis, Sequence

I assume that by sequence they mean ${\alpha\choose 1}$ $+{\alpha\choose 2}+...+{\alpha\choose n}$
I do not understand what should be $n$... Is it complex? Usual binomial coefficients have both $n$ and $\alpha$ integers and $n\leq\alpha$. In this problem can $n$ be more than $\alpha$??

2. Notice that they have "redefined" the numerator of the binomial coefficient as $(\alpha)(\alpha- 1)..$ rather than $\alpha !$ because factorial, is only defined for positive integers. Because they do use the notation "n!" in the denominator, n must be a non-negative integer.

You cannot even ask the question "can n be bigger than $\alpha$" because the complex number system is not an ordered field!

However, "sequence" does NOT mean a sum. That is a "series". They mean just the numbers $\left(\begin{array}{c}0 \\ \alpha\end{array}\right)$, $\left(\begin{array}{c}1 \\ \alpha\end{array}\right)$, ... themselves. Basically, the first thing you are asked to prove is that there is some real number M such that $\left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$ for all n.

3. Originally Posted by HallsofIvy
N
However, "sequence" does NOT mean a sum. That is a "series". They mean just the numbers $\left(\begin{array}{c}0 \\ \alpha\end{array}\right)$, $\left(\begin{array}{c}1 \\ \alpha\end{array}\right)$, ... themselves. Basically, the first thing you are asked to prove is that there is some real number M such that $\left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$ for all n.

you meant ${\alpha\choose\ n }< M$. I ask this because you wrote $\left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$

4. Originally Posted by andreas
you meant ${\alpha\choose\ n }< M$. I ask this because you wrote $\left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$
What HallsofIvy wrote is correct. When you want to show a sequence is bounded you have to show there is a positive number larger than the absolute value of each term. Therefore, you must include absolute values.

5. Originally Posted by ThePerfectHacker
What HallsofIvy wrote is correct. When you want to show a sequence is bounded you have to show there is a positive number larger than the absolute value of each term. Therefore, you must include absolute values.

I agree with that but why should $\alpha$ and $n$ exchange their places?

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I do not understand how to deal with inequalities where on one side is a complex number and on the other only real number... Can somebody explain intuitively what is going on?