This week in class, we covered one way to sum an arithmetico-geometric series. Now we're going to cover a different approach.
Let |r| < 1,
$$S = \sum_{k=0}^{\infty} r^k,$$and
$$T = \sum_{k=0}^{\infty} k r^k.$$Our approach is to write $T$ as a geometric series in terms of $S$ and $r$.
Give a closed form expression for $T$ in terms of $r$.
To find a closed form expression for T in terms of r, we can use the formula for the sum of an infinite geometric series and some algebraic manipulation.
Let's start by manipulating the expression for T:
T = \sum_{k=0}^{\infty} k r^k
We can rewrite the sum as:
T = 0 + 1r + 2r^2 + 3r^3 + ...
Now, let's multiply both sides of the equation by r:
rT = 0r + 1r^2 + 2r^3 + 3r^4 + ...
If we subtract the second equation from the first, we can eliminate all terms involving r:
T - rT = 0 + (1r - 0r) + (2r^2 - 1r^2) + (3r^3 - 2r^3) + ...
Simplifying the equation, we get:
T(1 - r) = 0 + r + r^2 + r^3 + ...
Now, this new sum on the right side of the equation looks familiar. It's the sum of an infinite geometric series with first term r and common ratio r. We can use the formula for the sum of an infinite geometric series to find its value.
The formula for the sum of an infinite geometric series is given by:
S = a / (1 - r)
Where:
- S is the sum of the series,
- a is the first term of the series, and
- r is the common ratio of the series.
In our case, the first term a is r, and the common ratio r is also r.
So, the right side of our equation simplifies to:
T(1 - r) = S = r / (1 - r)
Now, let's solve for T by dividing both sides of the equation by (1 - r):
T = (r / (1 - r)) / (1 - r)
Simplifying further, we get:
T = r / (1 - r)^2
Therefore, the closed form expression for T in terms of r is:
T = r / (1 - r)^2