convergence sequence

Question

let A₁ ≥A₂≥A_3........... ≥A_k ≥0 then 

lim_n →∝ [ A₁ⁿ + A₂ⁿ + ........+ A_kⁿ ]^1/n  is equal to

 

a) 1            b) 0          c) A_t             d) A_k              

ans is c  , please help me , solve it please

Answer 1

To be honest, I'm not even sure what A_t is in this context.  We can prove that the limit is between A₁ and A_k, which is perhaps the meaning of A_t.

 

Since A₁ is the largest entry, we have A₁ⁿ ≥ A_iⁿ, for 2 ≤ i ≤ k.

Likewise, since A_k is the smallest entry, we have A_kⁿ ≤ A_iⁿ, for i < k.

 

This leads to (A₁ⁿ + A₁ⁿ + ... + A₁ⁿ) ≥ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ) ≥ (A_kⁿ + A_kⁿ + ... + A_kⁿ),

where on the left and rightmost sides, A₁ⁿ and A_kⁿ are summed k times, respectively.  Continuing,

 

kA₁ⁿ ≥ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ) ≥ kA_kⁿ

(kA₁ⁿ)^1/n ≥ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ) ≥ (kA_kⁿ)^1/n

k^1/nA₁ ≥ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ)^1/n ≥ k^1/nA_k

 

Taking limits,

lim_n→∞ (k^1/nA₁) ≥ lim_n→∞ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ)^1/n ≥ lim_n→∞ (k^1/nA_k)

A₁*lim_n→∞(k^1/n) ≥ lim_n→∞ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ)^1/n ≥ A_k*lim_n→∞ (k^1/n)

A₁ ≥  lim_n→∞ (A₁ⁿ + A₂ⁿ + ... + A_kⁿ)^1/n ≥ A_k,

 

which shows that the limit is bounded by A₁ above and A_k below.