# If c <= (a_i)/(b_i) AA i, then c <= (sum_(i=1)^oo a_i)/(sum_(i=1)^oo b_i)

I'm trying to prove that the best moment tail bound is no worse than the best Chernoff bound. I have all pieces other than this little frustration here:
If $c\le \frac{{a}_{i}}{{b}_{i}}\mathrm{\forall }i$, then $c\le \frac{\sum _{i=1}^{\mathrm{\infty }}{a}_{i}}{\sum _{i=1}^{\mathrm{\infty }}{b}_{i}}$
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Sanaa Holder
Assuming all numbers are positive and that the series are convergent, you have
$c{b}_{i}\le {a}_{i}$
for all $i$. Hence, summing all up, you have
$c\sum _{i}{b}_{i}\le \sum _{i}{a}_{i}$
which is equivalent to
$c\le \frac{\sum _{i}{a}_{i}}{\sum _{i}{b}_{i}}$