LIMIT

is it
aritmetic mean of a_i ?

well arithmetic mean of a_i is wen m was not varying

here m is also tending to infinity

btw i dunno the ans

\text{SOLUTION : } \\ n\left(\sqrt[m]{\left( 1+\frac{a_1}{n}\right)\left( 1+\frac{a_2}{n}\right)\left( 1+\frac{a_3}{n}\right)\left( 1+\frac{a_4}{n}\right)........\left( 1+\frac{a_m}{n}\right)} -1\right)\\ n\left( \left(1+\left( \frac{\sum{a_i}}{n} +...........\right)\right)^{\frac{1}{m}}-1\right) \\ n\left(\left( \frac{\sum{a_i}}{mn}+.......) \right+ higher \ powers \right) \right) \\ \text{the constant terms comes out to be }\frac{\sum{a_i}}{m}=arithmetic \ mean \\ \lim_{m\rightarrow \infty} \frac{\sum{a_i}}{m} =0

@akari how u got ur last step ?

Σa_i is not constant its tending to ∞

so u cant hav lim as 0