Question 537595
{{{1+3/4+7/16+15/64+31/256}}} ......... to infinity
 
Consider the first term as {{{1/1}}}
 
{{{1/1+3/4+7/16+15/64+31/256}}} ......... to infinity
 
The numerators form the sequence 1,3,7,15,31 
 
That has nth term
 
{{{2^n-1}}}
 
The denominators form the sequence 1,4,16,64,256 
 
That has nth term
 
{{{2^(n-1)}}}
 
Therefore the sequence summed to infinity is
 
{{{sum(  ((2^n-1)/(2^(n-1))),n=1,infinity )}}} = {{{sum(  ((2^n)/(2^(n-1))),n=1,infinity )}}} - {{{sum(  (1/(2^(n-1))),n=1,infinity )}}} =
 
Subtract exponents in the first:
 
{{{sum(  (2^1),n=1,infinity )}}} - {{{sum(  (1/(2^(n-1))),n=1,infinity )}}}
 
The first series diverges to infinity and the second one is a geometric
series converging to 2, so the series diverges to infinity.
 
Edwin</pre>