A Converging Geometric Series

​
n
3
area = 0.875000000
​
This Demonstration shows that
∞
∑
n=1
n
1
2
=
1
2
+
1
4
+
1
8
+…+
n
1
2
=1
.
The square is half empty with area
1
2
when
n=1
. The square is
1
4
empty at
n=2
. In general, the square is
n
1
2
empty at step
n
and
1
2
+
1
4
+
1
8
+…+
n
1
2
full, which shows that
∞
∑
n=1
n
1
2
converges to 1.

Details

Snapshot 1: when
n=1
, the area is filled is
1
2
Snapshot 2: when
n=4
​
, a square with area
n
1
2
=
4
1
2
=
1
16
is added
Snapshot 3: where
n=25
and the area filled is very close to 1
The geometric series
∞
∑
n=1
n
r
converges if and only if
|r|<1
, and then the sum of the series is
r
1-r
. Convergence of
∞
∑
n=1
n
r
can be proven by the integral test, which states that if
f(x)
is continuous, decreasing, and positive, then
∞
∑
n=1
f(x)
converges if
∞
∫
1
f(x)dx
converges. In this case,
∞
∫
1
x
1
2
dx=
1
ln[4]
, so the integral converges and therefore the geometric series also converges.
Special thanks to the University of Illinois NetMath Program and the mathematics department at William Fremd High School.

References

[1] R. Bayer. "Proof By Picture." (Aug. 13, 2009) www.scribd.com/doc/254948592/Proof-by-Picture.
[2] M. Moody. "Convergence Tests for Infinite Series." (Jun 18, 2013) www.math.hmc.edu/calculus/tutorials/convergence.

External Links

Graphical Representation of Geometric Series
Visual Computation of Three Geometric Sums
Stepwise Trisection of a Square

Permanent Citation

Akane Hattori, Natsuki Okuda
​
​"A Converging Geometric Series"​
​http://demonstrations.wolfram.com/AConvergingGeometricSeries/​
​Wolfram Demonstrations Project​
​Published: June 19, 2013