a simple sum?

[Kai Schmidt]

Hello,

I'm sitting here with a nice little problem:

It is known that

t
----
\ k t
/ a * Binomial( t, k) = ( 1 + a )
----
k=0

This means that

t
----
-t \ k
(1 + a) * / a * Binomial( t, k) = 1
----
k=0

1
Now I suppose that adding only each m'th term gives - for large values of t:
m

t/m
----
-t \ k*m 1
limes (1 + a) * / a * Binomial( t, k*m) = -
t -> infinity ---- m
k=0

Is it possible to proof my suggestion, or is it even wrong?

Thanks, Kai Schmidt

__________________________________________________________________________/
TU Hamburg-Harburg Tel: +49 40 7718 3351
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Kai Schmidt email: K-Sc...@tu-harburg.d400.de
D - 21079 Hamburg

[Victor Miller]

Re: your simple sum:

f_m(a,t) = \sum_{i=0, m|i}^t a^k \binom{t}{k}

If a is real and >0 then, indeed,
\lim_{t \rightarrow \infty} (1+a)^{-t} f_m(a,t)
exists and equals 1/m. This is most easily seen as follows:

f_m(a,t) = (1/m) \sum{j=0}^{m-1} (1+\zeta^j a)^t

where \zeta = \exp(2 \pi i / m), a primitive m-th root of 1.

(Hint, expand each term on the right by the binomial theorem,
interchange summation and use the fact that
\sum{j=0}^{m-1} \zeta^{j k} = 0, except when m | k).

Finally, finish up by noticing that

|1+a \zeta^j|^2 = (1+a)^2 - 2 (1-cos(2 \pi j/m)) < (1+a)^2

so that all terms in limit, except the first, vanish.

--
Victor S. Miller -- vic...@ccr-p.ida.org
CCR, Princeton, NJ 08540

" ... Meanwhile, those of us who can compute can hardly be expected to keep
writing papers saying 'I can do the following useless calculation in 2
seconds', and indeed what editor would publish them?" -- Oliver Atkin