power series – A real variable solution to a problem posted on Terry Tao's blog

Here is my purely real variable solution.
We have this $ e ^ {x} $ and $ f (x) $ are both analytical, so their product $ phi (x) = e ^ {x} cdot f (x) $ is therefore $ phi (x) = overset { infty} { subset {n = 0} { sum}} c_n x ^ n $, or $ c_n = a_0 cdot frac {1} {(n-1)!} + … + frac {a_i} {i!} cdot frac {1} {(ni)!} + … + frac { a_n} {n!} $.
Since $ subset {x to infty} { lim} phi (x) = subset {x to infty} { lim} dfrac {f (x)} {e ^ {- x}} = C $, for each $ epsilon> $ 0 there is $ R> $ 0 and $ N in mathbb {N} $ such as:
$$ Bigg | overset {k} { subset {n = 0} { sum}} c_n x ^ n – C Bigg | leq Bigg | overset {k} { subset {n = 0} { sum}} c_n x ^ n – overset { infty} { subset {n = 0} { sum}} c_n x ^ n Bigg | + Bigg | overset { infty} { subset {n = 0} { sum}} c_n x ^ n – C Bigg | leq dfrac { epsilon} {2} + dfrac { epsilon} {2} = epsilon. $$

for everyone $ k> N $ and all $ x> R $. So,

$$ C- epsilon < overset{k}{underset{n=0}{sum}}c_n x^n N, x > R $$
That implies that,

$$ 0 = subset {x to infty} { lim} dfrac {C- epsilon} {x ^ k} <c_k N. $$
We deduce that $ phi (x) = overset {N} { subset {n = 0} { sum}} c_n x ^ n $ and because $ underet {x to infty} { lim} phi (x) = C $, $ phi (x) $ must be constant $ C. $ Therefore, $ f (x) = Ce ^ {- x} $, where the equations $ a_n = C (-1) ^ n $ follows.