SOLUTION: EVALUATE: {{{1+3/4+7/16+15/64+31/256}}} ......... to infinity

Algebra ->  Sequences-and-series -> SOLUTION: EVALUATE: {{{1+3/4+7/16+15/64+31/256}}} ......... to infinity      Log On


   

Question 537595: EVALUATE:
1%2B3%2F4%2B7%2F16%2B15%2F64%2B31%2F256 ......... to infinity
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20056) About Me  (Show Source):
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
1%2B3%2F4%2B7%2F16%2B15%2F64%2B31%2F256 ......... to infinity

Consider the first term as 1%2F1

1%2F1%2B3%2F4%2B7%2F16%2B15%2F64%2B31%2F256 ......... to infinity

The numerators form the sequence 1,3,7,15,31

That has nth term

2%5En-1

The denominators form the sequence 1,4,16,64,256

That has nth term

2%5E%28n-1%29

Therefore the sequence summed to infinity is

sum%28++%28%282%5En-1%29%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29 = sum%28++%28%282%5En%29%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29 - sum%28++%281%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29 =

Subtract exponents in the first:

sum%28++%282%5E1%29%2Cn=1%2Cinfinity+%29 - sum%28++%281%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29

The first series diverges to infinity and the second one is a geometric
series converging to 2, so the series diverges to infinity.

Edwin