\(M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2016}}\)

\(\Rightarrow\)\(2M=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}\)

\(\Rightarrow\)\(2M-M=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(\Rightarrow\)\(M=2-\frac{1}{2^{2016}}< 2\)

Vậy  M < 2 

Ta có : 

\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\) 

\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)

\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)

\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)

Đến chỗ này chịu!

Ta có

\(T=1+\frac{3}{1\cdot2^2}+\frac{4}{2\cdot2^2}+...+\frac{2017}{2^2\cdot2^{2014}}\) 

\(T=1+\frac{1}{2^2}\cdot\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\right)\)

2A= 2+2^3+2^4+...+2^2015+2^2016

2A-A=2^2016-1

A=(2^2016-1):2

Vì (2^2016-1):2 bé hơn 2^2016 nên A bé hơn 2^2016

\(M=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)

\(>1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1+\frac{1}{2}-\frac{1}{11}\)

\(>1+\frac{1}{2}-\frac{1}{6}=\frac{4}{3}\)