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