1      2       4        8          16         
   S = --- + ----- + ----- + ------- + ---------- + ..... 
        2      5       41     3,281    21,523,361              

What about letting a(n) be the following nth term

a(n) = 2^(1 + n)/(1 + 3^2^n) for n = 0, 1, 2, ...

Then the partial sums

s(n) = Sum (for k=0 to n) a(k)

is equal to 1/2, 9/10, 409/410 ....

with a limit as n tends to infinity of 1.

Edit with the benefit of earlier comments.

The proof of this is Bobs expression as corrected by Charlie for 1-s(n) = 2^(n+2)/(3^2^(n+1)-1)

Since 2^(n+2)/3^(n+2) = (2/3) ^ (n +2) tends to zero as n tends to infinity, it is sufficient to note that 3^2^(n+1)-1 > 3 ^ (n+2) for n >0  hence

0 <= 2^(n+2)/(3^2^(n+1)-1) < (2/3) ^ (n +2) ->0

= Lim (n -> infty) 1 - s(n)

From which it follows that Lim (n -> infty) s(n) = 1.

Edited on December 22, 2005, 6:14 am

Posted by goFish on 2005-12-22 03:58:28