Question

Sum of series to n terms 0.4+0.44+0.444+....

Answer 1

`0.4+0.44+0.444+\cdots=\frac4{10}+(44)/(100)+(444)/(1000)+\cdots=\displaystyle\sum_(k=1)^\infty(\frac49(10^k-1))/(10^k)=\frac49\sum_(k=1)^\infty\left(1-\frac1{10^k}\right)`

The
`n`th partial sum is given by


`S_n=\displaystyle\frac49\sum_(k=1)^n\left(1-\frac1{10^k}\right)`

`S_n\displaystyle=\frac49\bigg(\left(1-\frac1{10}\right)+\left(1-\frac1{100}\right)+\cdots+\left(1-\frac1{10^(n-1)}\right)+\left(1-\frac1{10^n}\right)\bigg)`

`S_n=\frac49\left(n-(10^n-1)/(9*10^n)\right)=\frac4{81}(9n-1+10^(-n))`

from which it follows that the infinite sum does not converge.