Determine convergence of divergence. If convergent, find the sum.
Sum (infinity, n = 0) (e/pi)^n
1 + .865 + .865^2 +.865^3 ......
this is a geometric series with the first term one and every successive term multiplied by r =.865 (approx)
If r <1
then the geometric series converges.
for the series
a + a r + a r^2 + a r^3 ----- a r^oo
the sum is
S = a/(1-r)
here a = 1
and r = .865
S = 1/(1-.865) = 7.4
Read:
http://en.wikipedia.org/wiki/Geometric_series
To determine the convergence of the series and find its sum, we can use the formula for the sum of an infinite geometric series. A geometric series has the general form:
S = a + ar + ar^2 + ar^3 + ...
where 'a' is the first term and 'r' is the common ratio.
In this case, our series is:
S = (e/pi)^0 + (e/pi)^1 + (e/pi)^2 + (e/pi)^3 + ...
We can rewrite this series as a geometric series by factoring out a common ratio:
S = 1 + (e/pi) + (e/pi)^2 + (e/pi)^3 + ...
Now we can compare this series to the general form of a geometric series:
S = a / (1 - r)
where |r| < 1 (to ensure convergence).
In our case, 'a' is the first term, which is 1, and 'r' is the common ratio, which is (e/pi).
Now, let's check the convergence condition |r| < 1:
|e/pi| < 1
Since the absolute value of e/pi is less than 1, this series meets the convergence condition.
Therefore, the series converges, and the sum can be found using the formula for the sum of a geometric series:
S = a / (1 - r)
S = 1 / (1 - e/pi)
So, the sum of the series (infinity, n = 0) (e/pi)^n is 1 / (1 - e/pi).