It is not
solution
Assuming the series converges it converges absolutely. Therefore
sum{n/2^(n-1)^ | n >= 1}
= sum{(n+1)/2^n^ | n >= 0}
= sum{n/2^n^ | n >= 0} + sum{1/2^n^ | n >= 0}
= sum{n/2^n^ | n >= 0} + 2
= sum{n/2^n^ | n >= 1} + 2
=>
sum{n/2^(n-1)^ | n >= 1} = sum{n/2^n^ | n >= 1} + 2
=>
2 = sum{n/2^(n-1)^ | n >= 1} - sum{n/2^n^ | n >= 1}
= sum{n/2^(n-1)^ - n/2^n^ | n >= 1}
= sum{n/2^n^ | n >= 1}
= 1/2 * sum{n/2^(n-1)^ | n >= 1}
=>
sum{n/2^(n-1)^ | n >= 1} = 4
