The Perturbation Method: Deriving Arithmetic-Geometric Series

Math · Series · Perturbation Method

Deriving an Arithmetic-Geometric Series with the Perturbation Method

A clean derivation of the closed form for \( \sum_{k=0}^{n} k2^k \) by expressing the same shifted sum in two different ways.

Target sum

We want a closed form for \( S_n = \sum_{k=0}^{n} k2^k \).

Main trick

Write \( S_{n+1} \) in two different forms, then set them equal.

Final result

\( S_n = (n-1)2^{n+1} + 2 \).

Introduction

We seek a closed-form expression for the arithmetic-geometric series \[ S_n = \sum_{k=0}^{n} k2^k. \]

The perturbation method works well here because the sum can be shifted and re-indexed in a way that exposes a second expression for the same object.

Main idea Compute \( S_{n+1} \) two different ways, then solve the resulting equation for \( S_n \).

Step 1

Define the Shifted Sum

First, isolate the last term:

\[ S_{n+1} = S_n + (n+1)2^{n+1}. \]

Next, re-index the same sum by isolating the first term and shifting the index:

$$ \begin{aligned} S_{n+1} &= \sum_{k=1}^{n+1} k2^k \\ &= \sum_{k=0}^{n} (k+1)2^{k+1} \\ &= 2 \sum_{k=0}^{n} (k+1)2^k \\ &= 2 \left( \sum_{k=0}^{n} k2^k + \sum_{k=0}^{n} 2^k \right) \\ &= 2(S_n + G_n), \end{aligned} $$

where \[ G_n = \sum_{k=0}^{n} 2^k \] is the associated geometric sub-sum.

Step 2

Solve the Geometric Sub-sum

Apply the same style of reasoning to \( G_n \):

$$ \begin{aligned} G_n + 2^{n+1} &= 1 + 2G_n \\ G_n &= 2^{n+1} - 1. \end{aligned} $$

This gives us the closed form we need to substitute back into the arithmetic-geometric relation.

Step 3

Finish the Derivation

Now equate the two expressions for \( S_{n+1} \):

\[ S_n + (n+1)2^{n+1} = 2S_n + 2G_n. \]

Substitute \( G_n = 2^{n+1} - 1 \):

\[ S_n + (n+1)2^{n+1} = 2S_n + 2(2^{n+1} - 1). \]

Rearranging gives:

\[ S_n = (n+1)2^{n+1} - 2^{n+2} + 2 \] \[ S_n = (n-1)2^{n+1} + 2. \]

Closed form \[ \sum_{k=0}^{n} k2^k = (n-1)2^{n+1} + 2. \]

Verification

Verification Table

n	Manual Summation	Value	Formula Result	Match
0	\( 0(2^0) \)	0	\( (-1)2^1 + 2 = 0 \)	✓
1	\( 0 + 1(2^1) \)	2	\( (0)2^2 + 2 = 2 \)	✓
2	\( 0 + 1(2^1) + 2(2^2) \)	10	\( (1)2^3 + 2 = 10 \)	✓
3	\( 0 + 1(2^1) + 2(2^2) + 3(2^3) \)	34	\( (2)2^4 + 2 = 34 \)	✓

FAQ

Frequently Asked Questions

These are the practical questions students usually have when first learning the perturbation method for arithmetic-geometric series.

Why do we use \( S_{n+1} \) instead of working directly with \( S_n \)?

Because shifting the sum by one index makes it easier to re-index the expression and create a second usable form of the same quantity.

Why introduce the geometric series \( G_n \)?

Because once the re-indexed form is expanded, the sum naturally splits into the original arithmetic-geometric part and a plain geometric series.

What makes this a perturbation method?

The method perturbs the original sum by shifting or modifying it slightly, then compares the new expression to the old one to isolate the closed form.

Is the final formula valid for \( n = 0 \)?

Yes. Substituting \( n = 0 \) gives \( (-1)2^1 + 2 = 0 \), which matches the original sum.

Why is the geometric sub-sum easier to solve?

Because geometric series have a standard closed form, and they respond very cleanly to the same shift-and-compare idea.

What is the simplest thing to remember from this derivation?

Write the shifted sum two ways, reduce the extra part to a geometric series, and solve by comparison.

Conclusion

Conclusion

The perturbation method is useful because it turns a weighted sum into something comparable to itself after a shift.

In this example, that shift exposed an ordinary geometric series, which made the rest of the algebra straightforward.

The final result is: \[ \sum_{k=0}^{n} k2^k = (n-1)2^{n+1} + 2. \]

Raell Dottin