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

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