MHB Find the sum of the first 11 terms of given series

[anemone]

Hi MHB,

This problem vexes me until my mind hurts.

Problem:

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

Attempt:

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

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

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

and I noticed this $$\sum_{k=1}^{11} \frac{4}{5(10(10^k)-1)} $$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:

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

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

Hi MHB,

This problem vexes me until my mind hurts.

Problem:

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

Attempt:

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

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

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.

 

[anemone]

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.