convergence

**alexandrabel90** · May 2nd 2010, 05:46 AM

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.

**Tikoloshe** · May 2nd 2010, 06:54 AM

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 $u_n=\frac{1}{n}$ unless $n$ contains the digit 4, then we can say that $\sum_{n=1}^\infty u_n=\sum_{n=1}^{9} u_n+\sum_{n=10}^{99} u_n+\sum_{n=100}^{999} u_n+\cdots$ , so if you let $s_m=\sum_{n\text{ has }m\text{ digits}}u_n$ , then we find that $\sum_{n=1}^\infty u_n=\sum_{m=1}^\infty s_m$ .

**alexandrabel90** · May 2nd 2010, 07:43 AM

i was reading the link that you posted but i cant understand the part about dividing it by 10^(r-1). may i know why we have to divide the number by that to get the sum? i thought the sum would just be 8(9^(r-1)).
thanks

**Tikoloshe** · May 2nd 2010, 09:38 AM

I think the point is that given natural numbers with $m$ digits, $8\times9^{m-1}$ of them have no 4’s, and each of their reciprocals is less than $10^{m-1}$ .

For example, if you consider all 2-digit numbers, the 9 numbers $14,\ldots,94$ must be removed as well as 9 more ( $40,41,42,43,45,\ldots,49$ ). 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

$s_2\leq72\times\frac{1}{10}$

It can be shown (this is your job) that in general, $s_m\leq8\left(\frac{9}{10}\right)^{m-1}$ .

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