Welcome to Part 18 of our wonderful 21 part of our Calculus Revisited Series. I hope you are enjoying this series. An announcement: Part 21 will be a "catch all" section - covering some of the topics we don't get to in detail in this series.
The next three blog entries will be about series. Today, the basics.
Series:
A series is a sum of terms in a sequence. The series can be finite or infinite. An infinite series is convergent (has a value) if the following is true:
∞
∑ a_n = S
n=i
and S < ∞
Here are some famous series:
Arithmetic Series:
a + (a + d) + (a + 2d) + (a + 3d) + ...
Sum for a finite arithmetic series:
n-1
∑ a + k * d
k=0
= a * n + (d * n * (n -1))/2
Geometric Series:
a + a * r + a * r^2 + a * r^3 + ...
Sum for an infinite geometric series, provided that |r| < 1:
∞
∑ a * r^k = a / (1 - r)
k = 0
Harmonic Series:
∞
∑ 1/n
n=1
= 1 + 1/2 + 1/3 + 1/4 + 1/5 + ....
There is no closed formula for a finite harmonic series. In addition the infinite harmonic series diverges (has no sum).
Series Properties:
∑ c * a(n) = c * ∑ a(n) (c is a constant)
∑ a(n) + b(n) = ∑ a(n) + ∑ b(n)
n
∑ k = n * (n + 1)/2
k=1
n
∑ k^2 = n * (n + 1) * (2n + 1)/6
k = 1
n
∑ k^3 = n^2 *(n + 1)^2 / 4
k = 1
Subtracting from a whole:
If t > 1:
k
∑ a(n) =
n = t
∑( a(n) for n = 1 to k) - ∑( a(n) for n = 1 to t - 1)
Problems
1. Calculate
6
∑ 50 - 2k
k = 0
This is an arithmetic series with n = 6 + 1 = 7, a = 50, and d = -2.
Then the sum is:
(50)(7) + (-2 * 7 * 6)/2 = 308
2. Show why the (infinite) harmonic series is divergent.
Here is one way:
Let S be the sum of the harmonic series, that is:
S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...
Note that:
1 + 1/2 > 2/2
1/3 + 1/4 > 2/4
1/5 + 1/6 > 2/6
1/7 + 1/8 > 2/8
and so on...
Then we have 2/2 + 2/4 + 2/6 + 2/8 + ....
Which simplifies to 1 + 1/2 + 1/3 + 1/4 + .... = S
Which implies S > S, which is impossible.
The harmonic series is divergent.
3. Find the sum:
∞
∑ .5^n + .3^n
n=0
We can first break the sum up:
∑(.5^n for n=0 to ∞) + ∑(.3^n for n=0 to ∞)
Both terms are geometric series. r = .5 for the first term and r = .3 for the second term.
Then:
∑(.5^n for n=0 to ∞) + ∑(.3^n for n=0 to ∞)
= 1 / (1 - .5) + 1 / (1 - .3)
= 1 / .5 - 1 / .7
= 24/7 ≈ 3.42857
4. Find the sum:
∞
∑ 2 / (5^n)
n = 1
This looks like a geometric series with r = 1/5. If we can put the sum into it's proper form, perhaps by adding and subtracting 2 / 5^0:
∑( 2 / (5^n) for n = 1 to ∞)
= ∑ ( 2 / (5^n) for n = 0 to ∞) - 2 / 5^0
= ∑ ( 2 / (5^n) for n = 0 to ∞) - 2
= 2 / ( 1 - 1/5) - 2
= 2 / (4/5) - 2
= 5/2 - 2 = 1/2
5. Find the sum:
20
∑ k^2
k=10
Again, looks like a k^2 series, but use the subtraction from the whole technique and:
∑( k^2 for k = 10 to 20)
= ∑(k^2 for k = 1 to 20) - ∑(k^2 for k = 1 to 10 - 1)
= (20)(20 + 1)(2 * 20 + 1)/6 - (9)(9 + 1)(2 * 9 + 1)/6
= 17220 / 6 + 1710 / 6
= 15510 / 6 = 2585
Thank you once again for joining us. Next time we will work some series convergence tests.
Eddie
This blog is property of Edward Shore. © 2012