Evaluating the Limit of a SeriesWhat is the limit of the series .9+.09+.009+...?

In summary, the conversation involved finding the sequence of partial sums and evaluating the limit of the series .9+.09+.009+..., as well as determining the sum of the series. Confusion arose over the correct sequence of partial sums and the limit, with some disagreement over the answer of the second part. Ultimately, it was suggested to recognize the series as .9999... and determine the simple number that it represents.
  • #1
GreenPrint
1,196
0

Homework Statement



Find the sequence of partial sums {S_n} and evaluate the limit of {S_n} for the following series

.9+.09+.009+...

What is .9+.09+.009+... equal to?

Homework Equations





The Attempt at a Solution



For the first part of the question (find the sequence of partial sums {S_n})
S_n=9(1/10)^n where n >= 1
my teachers assistant marked my answer correct

for the second part of the question (evaluate the limit of {S_n} for the following series .9+.09+.009+...)
I evaluated the limit of S_n by just simply taking the limit of S_n as n goes to infinity
lim n->inf S_n = 9*lim n->inf (1/10)^n = 0
My teachers assistant marked my question wrong and put
S_n = sigma[1,4] 9(1/10)^ character
I can't read what character he put
I don't see how this answer is correct and my answer is wrong. If my answer to finding S_n is correct then why can't I just evaluate the limit as n goes to infinity of S_n to "evaluate the limit of {S_n}? I don't understand what's wrong with my work.

for the third part (What is .9+.09+.009+... equal to?)
.9+.09+.009+... = sigma[n=1,inf] (1/10)^n = 9* (1/10)/(1-1/10) = 9* (1/10)/(9/10) = 9*1/10*10/9 = 1
my answer was marked correctly

I don't see how my answer to the second part is wrong. I hope somebody can clear up this confusion for me. Thanks for any help anyone can provide me.
 
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  • #2
I think your answer to the first part is also wrong. 9*(1/10)^n isn't a partial sum of the series. It's the nth term of the series.
 
  • #3
shouldn't the limit of the partial sums be the same as the sum of the series?
 
  • #4
Dick said:
I think your answer to the first part is also wrong. 9*(1/10)^n isn't a partial sum of the series. It's the nth term of the series.

I think your first part is okay. They didn't ask for the partial sum, they asked for the sequence of partial sums.

[EDIT] never mind, it does seem wrong.
 
  • #5
There are two different sequences here. The first is {.9, .09, .009, ..., 9 * 10-n, ...}
The sequence of partial sums is {.9, .99, .999, ...}
 
  • #6
Okay I found the sequence of partial sums to be [tex] S_n= \frac{9(1+10^{n-1})}{10^n} [/tex]

Then I found the limit of that to be 9/10.

Of course, this may easily be wrong. I'm just making an attempt.
 
  • #7
Actually, I think I MUST be wrong about 9/10, since isn't the limit of the sequence of partial sums supposed to equal to the sum of the series?

Oh, I see a problem with my partial sums. Oh well. Your turn GreenPrint. :)
 
  • #9
It might be much simpler to recognize that .9+ .09+ .009+ .0009+ ... is the same as .99999... where the "9" continues for ever. What very simple number is that?
 

1. What is the concept of an infinite series in Calculus 2?

An infinite series in Calculus 2 refers to a sum of an infinite number of terms. It is represented by sigma notation, where the index variable starts at a given value and goes to infinity. The terms in the series can either be finite or infinite.

2. How do you determine the convergence or divergence of an infinite series?

The convergence or divergence of an infinite series can be determined by using various tests such as the Comparison Test, Ratio Test, Root Test, or Integral Test. These tests examine the behavior of the terms in the series and determine whether they approach a finite value or go to infinity. If the terms approach a finite value, the series is said to be convergent. If the terms go to infinity, the series is said to be divergent.

3. What is the difference between a geometric series and a telescoping series?

A geometric series is a type of infinite series where each term is multiplied by a constant ratio. The sum of a geometric series can be found by using the formula Sn = a(1 - r^n)/(1 - r), where a is the first term and r is the common ratio. On the other hand, a telescoping series is a type of infinite series where most terms cancel each other out, leaving only a finite number of terms. The sum of a telescoping series can be found by simplifying the expression and evaluating the remaining terms.

4. Can an infinite series have a sum?

Yes, an infinite series can have a sum if it is convergent. If the series is divergent, it does not have a finite sum. The value of the sum can be found by adding the terms in the series together, either manually or using a calculator. However, it is important to note that some series have an infinite sum, meaning that the sum goes to infinity.

5. How is the concept of an infinite series used in real-life applications?

Infinite series is used in various real-life applications such as finance, physics, and engineering. For example, in finance, infinite series is used to calculate compound interest, where the interest is continuously compounded. In physics, infinite series is used to approximate complicated functions such as motion or heat transfer. In engineering, infinite series is used to design and analyze circuits, signal processing, and control systems.

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