2.4 - Bounding sums with integrals

2.4 - Bounding sums with integrals

A curve can tell you how large a sum is without adding all its terms. You will compare rectangles with integrals, derive finite bounds, and recognise sums whose shrinking rectangles approach an exact area.

Area comparisons

How can you estimate 1+1/2+⋯+1/n1+1/2+\cdots+1/n when nn is very large? Think of each term as the height of a rectangle of width 11.

On [1,2][1,2], the decreasing curve y=1/xy=1/x lies below the rectangle of height 11 and above the rectangle of height 1/21/2. On [2,3][2,3], the corresponding heights are 1/21/2 and 1/31/3. Adding these inequalities compares an area with sums.

[DIAGRAM: asset_name: Endpoint rectangles enclose an area; asset_slug: cam9231-integral-rectangles; caption: For a decreasing function, left endpoints give upper rectangles and right endpoints give lower rectangles; description: Two equal-scale panels show y equals 1 over x on 1 to 4 with three unit-width rectangles. The left panel uses heights 1, one half and one third above the curve; the right uses one half, one third and one quarter below it. Endpoint heights, width one and the upper or lower comparison are labelled.]
For a decreasing function, left endpoints give upper rectangles and right endpoints give lower rectangles

Upper and lower rectangle sums

On an interval divided into equal widths hh, a rectangle has area h×heighth\times\text{height}. For a nonnegative decreasing function, left-endpoint heights give an upper sum and right-endpoint heights a lower sum. For an increasing function, the roles reverse.

If the function is strictly increasing or strictly decreasing, each rectangle differs from the curve over an interval of positive width. The corresponding area inequalities are strict. For a function that can be constant, use non-strict inequalities instead.

Worked example 1: a decreasing curve

Question. Use unit-width rectangles to bound ∫131x dx\int_1^3\frac1x\,\mathrm{d}x.

Plan. Add right-endpoint heights for the lower bound and left-endpoint heights for the upper bound.

There are two intervals, [1,2][1,2] and [2,3][2,3]. Since 1/x1/x decreases,

12+13<∫131x dx<1+12.\frac12+\frac13<\int_1^3\frac1x\,\mathrm{d}x<1+\frac12.

Answer. 5/6<ln⁡3<3/25/6<\ln3<3/2.

Check. Direct evaluation gives ln⁡3=1.0986…\ln3=1.0986\ldots, which lies between 0.8333…0.8333\ldots and 1.51.5.

Change the monotonic direction. The endpoint rule must change with it.

Worked example 2: an increasing curve

Question. Use unit-width rectangles to bound ∫13x2 dx\int_1^3 x^2\,\mathrm{d}x.

Plan. Check that x2x^2 increases on this interval, then use left endpoints for the lower sum.

12+22<∫13x2 dx<22+325<∫13x2 dx<13\begin{aligned} 1^2+2^2&<\int_1^3x^2\,\mathrm{d}x<2^2+3^2 \\ 5&<\int_1^3x^2\,\mathrm{d}x<13 \end{aligned}

Each height has been multiplied by width 11.

Answer. The integral is strictly between 55 and 1313.

Check. Its exact value is [x3/3]13=26/3[x^3/3]_1^3=26/3, which satisfies both bounds.

For a curve that changes monotonic direction, split the interval at the turning point before choosing endpoint rectangles. A single left-endpoint rule need not bound the whole area.

Watch out

“Left endpoints always overestimate” is false. Decide whether the function increases or decreases on the interval before choosing the inequality sign.

Apply the increasing-curve rule to a new integrand.

Upper and lower bounds

To bound a sum, choose integration limits that make the rectangle heights match its first and last terms. Different placements can produce different valid bounds.

For a positive strictly decreasing function ff and an integer n≥2n\ge2, left rectangles on [1,n+1][1,n+1] have heights f(1),…,f(n)f(1),\ldots,f(n). Right rectangles on [1,n][1,n] have heights f(2),…,f(n)f(2),\ldots,f(n). Therefore

∫1n+1f(x) dx<∑r=1nf(r)<f(1)+∫1nf(x) dx.\int_1^{n+1}f(x)\,\mathrm{d}x <\sum_{r=1}^n f(r) <f(1)+\int_1^n f(x)\,\mathrm{d}x.

The isolated f(1)f(1) on the right repairs the missing first term. These comparisons come from rectangles; they are not formulae supplied in MF19.

Worked example 3: a harmonic sum

Question. Show that, for integers n≥2n\ge2,

ln⁡(n+1)<∑r=1n1r<1+ln⁡n.\ln(n+1)<\sum_{r=1}^n\frac1r<1+\ln n.

Plan. Use f(x)=1/xf(x)=1/x in the two rectangle placements.

Since f′(x)=−1/x2<0f'(x)=-1/x^2<0 for x>0x>0,

∫1n+11x dx<∑r=1n1r∑r=2n1r<∫1n1x dx\begin{aligned} \int_1^{n+1}\frac1x\,\mathrm{d}x&<\sum_{r=1}^n\frac1r \\ \sum_{r=2}^n\frac1r&<\int_1^n\frac1x\,\mathrm{d}x \end{aligned}

The first integral is ln⁡(n+1)\ln(n+1). Add 11 to the second inequality to include the term with r=1r=1.

Answer. ln⁡(n+1)<∑r=1n1/r<1+ln⁡n\ln(n+1)<\sum_{r=1}^n1/r<1+\ln n.

Check. At n=2n=2, the sum is 3/23/2, and ln⁡3<3/2<1+ln⁡2\ln3<3/2<1+\ln2. The general bounds also show why the sum cannot settle at a finite limit: its lower bound grows without bound.

Change only the exponent in the denominator. The geometric comparison stays the same; its integral changes.

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