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Find the sum of the first 11 terms of given series

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,677
Hi MHB,

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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Hi MHB,

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

\(\displaystyle \frac{19}{99}+\frac{199}{999}+\frac{1999}{9999}+ \cdots+11^{th} term=\sum_{k=1}^{11} \frac{2(10)^k-1}{10(10^k)-1}=\sum_{k=1}^{11} \left( \frac{1}{5}-\frac{4}{5(10(10^k)-1)} \right)\)
Alas, I don't think you made a valid move there. It is perilous to cancel things that aren't factors. You can separate out the two sums:
$$ \sum_{k=1}^{11} \frac{2(10)^k-1}{10^{k+1}-1}=
2\sum_{k=1}^{11} \frac{10^k}{10^{k+1}-1}
- \sum_{k=1}^{11} \frac{1}{10^{k+1}-1}.$$
I'm not sure where this lands you. I would just hand it over to a CAS at this point.

You could try get a common denominator, which would look something like
$$\prod_{j=1}^{11}(10^{j+1}-1).$$
The resulting expression would be rather tedious to sort out.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,677
Alas, I don't think you made a valid move there. It is perilous to cancel things that aren't factors. You can separate out the two sums:
$$ \sum_{k=1}^{11} \frac{2(10)^k-1}{10^{k+1}-1}=
2\sum_{k=1}^{11} \frac{10^k}{10^{k+1}-1}
- \sum_{k=1}^{11} \frac{1}{10^{k+1}-1}.$$
I'm not sure where this lands you. I would just hand it over to a CAS at this point.

You could try get a common denominator, which would look something like
$$\prod_{j=1}^{11}(10^{j+1}-1).$$
The resulting expression would be rather tedious to sort out.
Thank you for the reply, Ackbach!:)

Yes, after spending more time to attempt to overcome the problem on my own, I agree that it's best to let the problem be handled by wolfram or any other CAS in order to get the exact value of the sum.