\(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+.....+\frac{n}{2^n}+......+\frac{2017}{2^{2017}}\)
Với n > 2 thì \(\frac{n}{2^n}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}\)
\(\frac{n+1}{2^{n-1}}=\frac{n+1}{2^n:2}=\frac{n+1}{\frac{2^n}{2}}=\frac{2^{\left(n+1\right)}}{2^n}\)
\(\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}=\frac{2^{n+2}}{2^n}-\frac{n+2}{2^n}\)
\(=\frac{2^{n+2}-n-2}{2^n}\)
\(=\frac{n}{2^n}\)
\(\Leftrightarrow S=\frac{1}{2}+\left(\frac{2+1}{2^{2-1}}-\frac{2+2}{2^2}\right)+.....+\frac{2016+1}{2^{2015}}-\frac{2018}{2^{2016}}\)
\(=\frac{2017+1}{2^{2016}}-\frac{2019}{2^{2017}}\)
\(S=\frac{1}{2}+\frac{3}{2}-\frac{2019}{2017}\)
\(S=2-\frac{2019}{2017}\)
\(\Leftrightarrow S=2-\frac{2019}{2017}< 2\)
Hay \(S< 2\)
