Find the sum of the first 11 terms of given series

In summary, the problem is to find the sum of the first 11 terms of a given series. The attempt to find the expression of the nth term led to a dead end, as it was not a geometric series. An alternative approach was attempted but also proved to be a cul-de-sac. It is suggested to use a computer algebra system to get the exact value of the sum.
  • #1
anemone
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MHB
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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.
 
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  • #2
anemone said:
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.
 
  • #3
Ackbach said:
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.
 

FAQ: Find the sum of the first 11 terms of given series

1. What is the formula for finding the sum of the first 11 terms of a series?

The formula for finding the sum of the first 11 terms of a series is:
Sum = (n/2)(2a + (n-1)d)

2. How do you determine the value of 'n' in the formula?

'n' represents the number of terms in the series. In this case, since we are finding the sum of the first 11 terms, 'n' would be equal to 11.

3. What does 'a' and 'd' stand for in the formula?

'a' represents the first term in the series and 'd' represents the common difference between each term in the series.

4. Can you provide an example of finding the sum of the first 11 terms of a series?

For example, if the series is 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, then 'a' would be 2 and 'd' would be 3. Plugging in these values into the formula, we get:
Sum = (11/2)(2(2) + (11-1)3)
Sum = (11/2)(4 + 30)
Sum = (11/2)(34)
Sum = 187

5. Is there an easier way to find the sum of the first 11 terms of a series?

Yes, you can also use the formula:
Sum = (n/2)(a + l)
Where 'l' represents the last term in the series. This formula is especially useful if the series follows a pattern, making it easier to find the last term.

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