Assuming the series converges it converges absolutely. Therefore

sumn/2^(n-1) n >= 1
= sum(n+1)/2ⁿ n >= 0
= sumn/2ⁿ n >= 0 + sum1/2ⁿ n >= 0
= sumn/2ⁿ n >= 0 + 2
= sumn/2ⁿ n >= 1 + 2

=>

sumn/2^(n-1) n >= 1 = sumn/2ⁿ n >= 1 + 2

=>

2 = sumn/2^(n-1) n >= 1 - sumn/2ⁿ n >= 1
= sumn/2^(n-1) - n/2ⁿ n >= 1
= sumn/2ⁿ n >= 1
= 1/2 * sumn/2^(n-1) n >= 1

=>

sumn/2^(n-1) n >= 1 = 4

It is not a telescoping series

https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd solutions/2024-05-18_not-a-telescoping-series.html