Integrating Infinite Series Involving Polynomials: A Detailed Analysis

When dealing with mathematical series, it's often crucial to understand how to manipulate and integrate them, particularly when these series involve polynomials. In this article, we'll explore the integration of a specific type of series, 1 2x 3x^2 4x^3 ldots, using calculus techniques. This series can be expressed as a function and then integrated, leading to powerful insights into its behavior and properties.

Introduction to the Series

The series in question is defined as:

1 2x 3x^2 4x^3 ldots sum_{n1}^{infty} nx^{n-1}

This series can be derived from a known geometric series by differentiating it. We'll first review the geometric series and then proceed to derive and integrate our series.

The Geometric Series

The geometric series is given by:

1 x x^2 x^3 ldots sum_{n0}^{infty} x^n frac{1}{1-x}

This series converges for |x| . To manipulate our original series, we differentiate this geometric series with respect to x.

Differentiating the Geometric Series

Differentiating both sides of the geometric series:

frac{d}{dx} left( sum_{n0}^{infty} x^n right) frac{d}{dx} left( frac{1}{1-x} right)

The left-hand side becomes:

sum_{n1}^{infty} nx^{n-1}

The right-hand side becomes:

frac{d}{dx} left( frac{1}{1-x} right) frac{1}{(1-x)^2}

Thus, we have:

1 2x 3x^2 4x^3 ldots frac{1}{(1-x)^2}

Integration of the Series

To integrate the series:

1 2x 3x^2 4x^3 ldots frac{1}{(1-x)^2}

We need to integrate:

int frac{1}{(1-x)^2} , dx

We can solve this by using the power rule for integration:

int u^{-2} , du -frac{1}{u} C

Let u 1 - x, then du -dx. Substituting these into the integral, we get:

int frac{1}{(1-x)^2} , dx -frac{1}{1-x} C

Conclusion

The integral of the series is given by:

-frac{1}{1-x} C

This solution is valid for |x| . As a result, we have successfully integrated the infinite series involving polynomials.

Additional Insight: Differentiation of the Series

Let's consider the differentiation of the series for a moment. If we denote:

y 1 2x 3x^2 4x^3 ldots

Differentiating y with respect to x gives:

frac{dy}{dx} 2 6x 12x^2 24x^3 ldots

This can be written as:

frac{dy}{dx} 2 cdot frac{1!}{0!}x^0 3 cdot frac{2!}{1!}x^1 4 cdot frac{3!}{2!}x^2 5 cdot frac{4!}{3!}x^3 ldots

Or more simply as:

frac{dy}{dx} sum_{n1}^{infty} n cdot frac{n!}{(n-1)!}x^{n-1}

The integral of the series can now be linked back to the differentiation, providing a deeper understanding of the relationship between integration and differentiation.

Simplified Integral Sum

Considering the integral sum up to n terms can be left at:

int 1 2x 3x^2 4x^3 ldots dx int frac{x^{n-1}}{1-x^n} , dx

This can be simplified further using the geometric series sum formula:

int x^0 x^1 x^2 x^3 ldots x^{n-1} dx frac{x^n - 1}{n(x-1)}

Thus, the simplified integral up to n terms is:

frac{xx^n - 1}{n(x-1)}

This helps in understanding the behavior of the series and its integral over a finite interval.