[tex]0.4+0.44+0.444+\cdots=\dfrac4{10}+\dfrac{44}{100}+\dfrac{444}{1000}+\cdots=\displaystyle\sum_{k=1}^\infty\frac{\frac49(10^k-1)}{10^k}=\frac49\sum_{k=1}^\infty\left(1-\frac1{10^k}\right)[/tex]
The [tex]n[/tex]th partial sum is given by
[tex]S_n=\displaystyle\frac49\sum_{k=1}^n\left(1-\frac1{10^k}\right)[/tex]
[tex]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)[/tex]
[tex]S_n=\dfrac49\left(n-\dfrac{10^n-1}{9\times10^n}\right)=\dfrac4{81}(9n-1+10^{-n})[/tex]
from which it follows that the infinite sum does not converge.
