Consider the series
∑ ∞ n=1
(13/10^n)
Determine whether the series converges, and if it converges, determine its value.
Converges (y/n):
Yippee ki-yay! Let's see if this series converges or not. We have ∑ ∞ n=1 (13/10^n). Well, well, well, it's a geometric series with a common ratio of 1/10. Now, my funny friend, for a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, 1/10 is definitely less than 1, so we're on the right track.
Now, let's calculate the sum of this series (if it converges). The formula for the sum of an infinite geometric series is S = a / (1 - r), where "a" is the first term and "r" is the common ratio.
In our case, the first term "a" is 13, and the common ratio "r" is 1/10. Plug those values in, and we get S = 13 / (1 - 1/10).
Doing some math wizardry, we simplify it to 13 / (9/10), which turns out to be 130/9.
So, drumroll, please! The series converges, my friend, and its value is 130/9.
Hope I put a big smile on your face with my clown bot calculations! 🤡
To determine whether the series ∑ ∞ n=1 (13/10^n) converges, we can use the formula for the sum of a geometric series.
The formula for the sum of an infinite geometric series is:
S = a / (1 - r)
Where "S" is the sum of the series, "a" is the first term, and "r" is the common ratio.
In this case, the first term "a" is 13/10^1 = 13/10, and the common ratio "r" is 10/10 = 1.
Plugging these values into the formula, we get:
S = (13/10) / (1 - 1)
Since the denominator 1 - 1 = 0, the sum of the series does not exist (dividing by zero is undefined).
Therefore, the series ∑ ∞ n=1 (13/10^n) does not converge.
To determine whether the series converges, we can use the concept of geometric series. A geometric series is a series in which each term is a constant multiple of the previous term.
The given series can be written as:
∑ ∞ n=1 (13/10^n) = 13/10^1 + 13/10^2 + 13/10^3 + ...
We can see that the common ratio here is 1/10. In order for the series to converge, the absolute value of the common ratio must be less than 1.
|1/10| = 1/10 < 1
Since the absolute value of the common ratio is less than 1, the series converges.
Now, to find its value, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
Where S is the sum, a is the first term of the series, and r is the common ratio.
In this case, the first term a is 13/10^1 = 13/10, and the common ratio r is 1/10.
S = (13/10) / (1 - 1/10)
= (13/10) / (9/10)
= 13/9
Therefore, the value of the series is 13/9.
Converges (y/n): Yes
Value of the series: 13/9
The 13 is just a scale factor, and has no bearing.
You know that
∞
∑ 1/10^n = 0.11111111 = 1/9
n=1