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- Feb 14, 2012

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This problem vexes me until my mind hurts.

Problem:

Find the sum of the first 11 terms of the series \(\displaystyle \frac{19}{99}+\frac{199}{999}+\frac{1999}{9999}+ \cdots\)

Attempt:

I managed only to find the expression of the nth term of the given series and I got

\(\displaystyle \frac{19}{99}+\frac{199}{999}+\frac{1999}{9999}+ \cdots+11^{th} term=\sum_{k=1}^{11} \frac{2(10)^k-1}{10(10^k)-1}=\sum_{k=1}^{11} \left( \frac{1}{5}-\frac{4}{5(10(10^k)-1)} \right)\)

\(\displaystyle =\frac{11}{5}-\frac{4}{5} \left( \frac{1}{99}+\frac{1}{999}+ \cdots + 11^{th} term \right)\)

and I noticed this \(\displaystyle \sum_{k=1}^{11} \frac{4}{5(10(10^k)-1)} \)isn't a geometric series and thus by rewriting the problem in this manner is a dead end and it won't solve the problem.

So I tried to break the given series as follows:

\(\displaystyle \frac{19}{99}+\frac{199}{999}+\frac{1999}{9999}+ \cdots+11^{th} term=(\frac{1}{99}+\frac{9}{99})+(\frac{1}{999}+ \frac{2(99)}{999})+ \cdots+11th term\)

\(\displaystyle =\frac{1}{99}+\frac{1}{999}+ \cdots + 11^{th} term+2 \left( \frac{99}{999}+\frac{999}{9999}+ \cdots + 11^{th} term \right)\)

and I know this is another cul-de-sac and I am getting so mad right now, I just don't see how to approach the problem.

Could anyone help me, please?

Thanks in advance.