Verified answer

You can deal with the base case.

If it holds for 1, 2, ... n, then

Then,

p_(n+1) <= p_1 p_2 .. p_n + 1

-----------<= 2^(2^0) 2^(2^1) ... 2^(2^(n-1)) + 1 (using the hypothesis)

--------------= 2^(2^0 + 2^1 + ... 2^(n-1) ) + 1

..............= 2^(2^n - 1) + 1 <= 2^(2^n - 1) + 2^(2^n - 1) = 2* 2^(2^n - 1) = 2^(2^n - 1 + 1) = 2^(2^n).

So it also is true for n + 1. Check it, since its a messy sum and I may have made a mistake typing it.

attempt comparing it to the sequence (2n +3 )/n^3 = (2/n^2 + 3/n^3) sum of two convergent p-sequence. because 0 is below or equivalent to |(2n+3)/((n^3)+a million)| it is below or equivalent to | (2n +3 )/n^3| as n techniques infinity, then by skill of assessment with the convergent p-sequence, your sequence additionally converges.