Determine the generating function for the sequence of natural logarithms of successive natural numbers | Step-by-Step Solution

Finding the Generating Function for Natural Logarithms

1. What We're Solving:

We want to find the generating function A(x) = ∑(n=0 to ∞) log(n+1) · x^n, which encodes the sequence {log(1), log(2), log(3), log(4), ...} as coefficients of a power series.

2. The Approach:

This is a beautiful problem that connects generating functions with calculus! Instead of trying to find a closed form directly, we'll use a clever technique: we'll relate our generating function to simpler ones we already know, then use differentiation and integration to transform between them. The key insight is recognizing that logarithms are related to derivatives of power functions.

3. Step-by-Step Solution:

Step 1: Start with a familiar generating function Let's begin with the geometric series, which you probably know well: $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$$

Step 2: Create a connection to our target We want coefficients that look like log(n+1). Notice that: $$\frac{d}{dx}\left[x^{n+1}\right] = (n+1)x^n$$

This suggests we should look at derivatives! But we need logarithms, not just (n+1).

Step 3: Use the key insight about logarithms Here's the crucial observation: We know that ∫(1/t)dt = log(t). Let's use this!

If we integrate our geometric series term by term from 0 to x: $$\int_0^x \frac{1}{1-t} dt = \int_0^x \sum_{n=0}^{\infty} t^n dt = \sum_{n=0}^{\infty} \int_0^x t^n dt$$

Step 4: Evaluate the integrals $$\int_0^x t^n dt = \frac{x^{n+1}}{n+1}$$

So: $$\int_0^x \frac{1}{1-t} dt = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$$

Step 5: Evaluate the left side $$\int_0^x \frac{1}{1-t} dt = -\log(1-x)$$

Therefore: $$-\log(1-x) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{x^n}{n}$$

Step 6: Transform to get our desired form Now we have a generating function where the coefficients are 1/n, but we want log(n+1). Let's differentiate both sides:

$$\frac{d}{dx}[-\log(1-x)] = \frac{d}{dx}\left[\sum_{n=1}^{\infty} \frac{x^n}{n}\right]$$

$$\frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1} = \sum_{n=0}^{\infty} x^n$$

Step 7: Apply the operator that gives us logarithms We can write: $$A(x) = \sum_{n=0}^{\infty} \log(n+1) x^n$$

Using the integral representation: log(n+1) = ∫₀¹ (1-t^(n+1))/(1-t) dt, we get: $$A(x) = \int_0^1 \frac{1-1/(1-xt)}{(1-t)} dt = \int_0^1 \frac{1}{(1-t)(1-xt)} dt$$

4. The Answer:

The generating function is: $$A(x) = \frac{-\log(1-x)}{x}$$

You can verify this by noting that the coefficient of x^n in the power series expansion of -log(1-x)/x is indeed log(n+1).

5. Memory Tip:

Remember the pattern: "Logarithms come from integrating 1/t, and generating functions often involve geometric series." When you see logarithmic coefficients, think about connecting to -log(1-x) through integration or differentiation operations!

The beauty here is how we transformed a simple geometric series into something that captures the logarithmic sequence. This technique of relating unknown generating functions to known ones through calculus operations is incredibly powerful!