let the summation form n= 1 to infinity for Un be the series obtains from summation from n=1 to infinity for 1/n by deleting all the i/n terms where n contains a digit equal to 4. prove that the series converges.
This is a variant of the Kempner series. To show it converges, try to find a bound on partial sums containing m digits.
That is, if $\displaystyle u_n=\frac{1}{n}$ unless $\displaystyle n$ contains the digit 4, then we can say that $\displaystyle \sum_{n=1}^\infty u_n=\sum_{n=1}^{9} u_n+\sum_{n=10}^{99} u_n+\sum_{n=100}^{999} u_n+\cdots$, so if you let $\displaystyle s_m=\sum_{n\text{ has }m\text{ digits}}u_n$, then we find that $\displaystyle \sum_{n=1}^\infty u_n=\sum_{m=1}^\infty s_m$.
I think the point is that given natural numbers with $\displaystyle m$ digits, $\displaystyle 8\times9^{m-1}$ of them have no 4’s, and each of their reciprocals is less than $\displaystyle 10^{m-1}$.
For example, if you consider all 2-digit numbers, the 9 numbers $\displaystyle 14,\ldots,94$ must be removed as well as 9 more ($\displaystyle 40,41,42,43,45,\ldots,49$). Since there are a total of 90 numbers from 10 to 99, and 18 of them contain the digit 4, only 72 of them need to be considered. Moreover, each of these has a reciprocal less than the reciprocal of the least (which is 1/10). Thus, using the notation of my previous post, we find that
$\displaystyle s_2\leq72\times\frac{1}{10}$
It can be shown (this is your job) that in general, $\displaystyle s_m\leq8\left(\frac{9}{10}\right)^{m-1}$.