What Is the Limit of the Sequence Defined by the Sum of Binomial Coefficients?

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The discussion focuses on finding the limit of the series defined by the sum of binomial coefficients, specifically ƩK(n+m,n)zn. The Cauchy root test is applied to determine the convergence of the series, with the limit expressed as 1/R = limn->∞[(K(n+m,n))½]. Participants clarify that the series is infinite, running from n=0 to infinity, and express confusion about the convergence value. A suggestion is made to substitute z with -t, which transforms the coefficients into negative binomial coefficients, potentially simplifying the evaluation of the sum. The conversation emphasizes the need for clarity in defining the series and its parameters for effective analysis.
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Homework Statement


I want to find the limit of ƩK(n+m,n)zn
K(a,b) being the binomial coefficient.

Homework Equations


Cauchy root test?

The Attempt at a Solution



Trying the cauchy root test I get:

1/R = limn->∞[(K(n+m,n))½]

But what do I do from here?
 
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The "Cauchy root test" tells you whether or not a series converges. It says nothing about what it converges to. If I read this correctly, you have
\sum \begin{pmatrix}n+m \\ n\end{pmatrix}z^n

The sum is over n with m fixed? And it is a finite sum? n goes from 0 to what?
 
well maybe I named it wrong, but I meant the formula stated above, which gives an explicit expression for the radius of convergence, R.
And the sum is from zero to infinity. Sorry for the lack of information :)
 
aaaa202 said:
well maybe I named it wrong, but I meant the formula stated above, which gives an explicit expression for the radius of convergence, R.
And the sum is from zero to infinity. Sorry for the lack of information :)

If you set z = -t, the coefficient of t^n is the "negative binomial" coefficient:
(-1)^n {n+m \choose n} = {-m \choose n}. That should allow you to evaluate the sum explicitly.

RGV
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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