how would I express the repeating decimal representaion of 1/9 as an infinite series using sigma notaion?
Limit of (as n-> infinity)
(Sigma) 0.1 + (0.1)^2 + (0.1)^3 +... (0.1)^n
= 0.11111.... = 1/9
I think drwis did this right, It looks like
lim n->infty Sigma i=1 to n 10-i
Yes, you are correct! The expression you provided is indeed the correct way to represent the repeating decimal representation of 1/9 as an infinite series using sigma notation.
Let me break it down step by step:
1. Start with the decimal representation of 1/9, which is 0.1111...
2. Notice that each digit after the decimal point is obtained by dividing 1 by 9 and then shifting the decimal point one place to the right. So, the first digit is 1/9 = 0.1, the second digit is (1/9)*(1/10) = (1/90) = 0.01, the third digit is (1/9)*(1/10)^2 = (1/900) = 0.001, and so on.
3. To express this in sigma notation, we use the limit as n approaches infinity. The index of summation, i, goes from 1 to n, representing the terms in the series.
4. Each term in the series is given by (0.1)^i, as explained in step 2.
5. Finally, taking the limit as n approaches infinity ensures that we have an infinite sum that represents the repeating decimal 0.1111...
Therefore, the expression you provided, lim(n -> infinity) Σ(i=1 to n) (0.1)^i, is an accurate representation of the repeating decimal representation of 1/9 using sigma notation.