Evaluating Polynomial Riemann Limits with Sum Formulas

From a Riemann Sum to a Limit

A Riemann sum for a function on an interval is built by dividing the interval into subintervals, choosing a sample point in each one, and adding the function value at that point times the subinterval width. When the sample points depend on an integer \(n\), evaluating the limit often becomes an algebra problem: express each sample point in terms of its index, substitute into the function, and use finite-sum formulas.

This method is useful because a sum that initially contains many terms can be rewritten as a formula involving \(n\). The limit as \(n\) increases can then be found by simplifying that formula and examining its highest powers of \(n\).

Here \(n\) is the number of subintervals and \(k\) is the index identifying one of them. Keeping their roles distinct is essential: \(n\) changes as the partition is refined, while \(k\) runs through the terms of a particular sum.

Finite-Sum Formulas You Need

For the sums in this tutorial, the basic formulas are:

The first formula counts \(n\) terms, each equal to \(1\). The others let us replace a sum of powers of the index with an expression in \(n\). These are finite identities, valid for every positive integer \(n\), not approximations.

Why the Sum-of-Squares Formula Works

A short derivation shows how such formulas can be established. Expand the difference between consecutive cubes:

Sum both sides from \(k=1\) to \(k=n\). On the left, consecutive terms cancel:

On the right, use the sum formulas for \(k\) and \(1\):

Rearrange to isolate the sum of squares:

Dividing by \(3\) gives \(\sum_{k=1}^{n}k^2=n(n+1)(2n+1)/6\). This derivation also illustrates why a power sum is helpful: after summing a telescoping difference, the desired finite sum can be isolated using simpler formulas.

A Reliable Evaluation Process

Worked Examples

Why the Algebra Gives the Riemann-Sum Limit

The calculation is more than a collection of substitutions. For each fixed \(n\), the sample points and the width determine an exact finite sum. The sum formulas rewrite that finite sum exactly as a function of \(n\). Taking its limit therefore evaluates the original sequence of Riemann sums, rather than an approximation to that sequence.

For example, in the quadratic case, the sample points are \(2k/n\), and each rectangle has width \(2/n\). The resulting sum is exactly \(8\sum k^2/n^3\). Since the sum-of-squares identity is valid for every positive integer \(n\), replacing the summation by its formula changes nothing. The remaining limit follows from ordinary algebra. By the Riemann-sum definition of a definite integral, when the function is continuous on the interval, this limit is also the definite integral over that interval.

Common Mistake: Losing a Factor of the Width

A frequent error is to substitute the sample point into \(f\) but forget to multiply by \(\Delta x\). For the quadratic example, an incorrect calculation might be

This is not the Riemann sum: it adds function values but omits the rectangle width \(2/n\). The correct term is

Without the width, the sum grows on the scale of \(n\), rather than approaching a finite area. A second indexing error is to use \(k/n\) as the right endpoint on \([a,b]\) without including \(a\) or scaling by \(b-a\). Always derive \(x_k=a+k(b-a)/n\) from the interval before substituting into \(f\).

How to Recognize a Successful Calculation

After applying the finite-sum formulas, a polynomial Riemann sum usually becomes a rational expression in \(n\). A finite limit is obtained when the powers of \(n\) in the numerator and denominator balance so the expression can be simplified to a constant plus terms that approach zero. If the resulting expression instead grows without bound, check whether the width was included, whether the sample point matches the stated interval, and whether the sum formula was applied with the correct powers.

For a polynomial of higher degree, expand \(f(x_k)\) into terms involving powers of \(k\) and \(n\), then use the corresponding finite-sum formula for each power. The same structure remains: sample point, function value, width, finite sum, simplification, limit. Careful bookkeeping matters more than speed.