Giải thích các bước giải:

Ta có:

$T=1+\dfrac22+\dfrac3{2^2}+\dfrac4{2^3}+...+\dfrac{2019}{2^{2018}}$

$\to 2T=2+\dfrac21+\dfrac3{2}+\dfrac4{2^2}+...+\dfrac{2019}{2^{2017}}$

$\to 2T-T=2+\dfrac11+\dfrac12+\dfrac1{2^2}+...+\dfrac1{2^{2017}}-\dfrac{2019}{2^{2018}}$

$\to T=3+\dfrac12+\dfrac1{2^2}+...+\dfrac1{2^{2017}}-\dfrac{2019}{2^{2018}}$

Đặt $A=\dfrac12+\dfrac1{2^2}+...+\dfrac1{2^{2017}}$

$\to 2A=1+\dfrac12+...+\dfrac1{2^{2016}}$

$\to 2A-A=1-\dfrac1{2^{2017}}$

$\to A=1-\dfrac1{2^{2017}}$

$\to T=3+1-\dfrac1{2^{2017}}-\dfrac{2019}{2^{2018}}$

$\to T>3$