Mostrar que 3 ^ (1/3) xx9 ^ (1/9) xx27 ^ (1/27) ... a l'infinit = 3 ^ (3/4). Com?

Resposta:

Mirar abaix.

Explicació:

`3 ^ (1/3) xx9 ^ (1/9) xx27 ^ (1/27) cdots = 3 ^ (1/3) xx 3 ^ (2/9) xx 3 ^ (3/27) cdots = 3 ^ (1/3 + 2/9 + 3/27 + cdots + n / 3 ^ n + cdots) = 3 ^ S`

amb

`S = sum_ (k = 1) ^ oo n / 3 ^ n =?`

Ho sabem `sum_ (k = 1) ^ oo k x ^ k = x d / (dx) suma_ (k = 1) ^ oo x ^ k`

i també per a `abs x <1`

`sum_ (k = 1) ^ oo x ^ k = 1 / (1-x) -1` i `d / (dx) (1 / (1-x) -1) = 1 / (1-x) ^ 2` llavors

`sum_ (k = 1) ^ oo k x ^ k = x / (1-x) ^ 2` i per `x = 1/3` tenim

`S = 3/4` llavors finalment

`3 ^ S = 3 ^ (3/4)`