Bound a remaining tail
How much is left after ten terms?
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}\) converges, and its sum is \(s\). The first 10 terms add to \(S_{10}\approx1.082037\).
The tail \(R_{10}=s-S_{10}\) is the part still missing, the terms from \(\frac1{11^4}\) on. Each teal rectangle is one unit wide and as tall as its term, so its area equals the term. Its orange corner is on the curve at \(x=11\), 12, 13 and so on.
With the first button, each rectangle lies to the left of its corner, so the first one starts at \(x=10\). The curve falls, so the rectangles lie under it and the tail is at most the area from 10 to infinity, \(\frac1{3000}\).
With the second button, each rectangle moves to the right of its corner, so the first one starts at \(x=11\). They lie above the curve, so the tail is at least the area from 11 to infinity, \(\frac1{3993}\).
Add \(S_{10}\) to both bounds. The sum \(s\) lies between 1.08229 and 1.08237.
The two bounds start one unit apart
The tail after \(n\) terms starts with the term \(a_{n+1}\), whose rectangle is outlined. In both positions its orange corner stays on the curve at \(x=n+1\).
Under the curve, the rectangle ends at \(x=n+1\), so it starts at \(x=n\) and the tail is at most the area from \(n\) to infinity.
Above the curve, the rectangle starts at \(x=n+1\), so the tail is at least the area from \(n+1\) to infinity.
The lower bound therefore starts one unit later than the upper bound.
Notation
The tail after \(n\) terms is what the partial sum leaves out.
\[R_n=s-S_n=a_{n+1}+a_{n+2}+\cdots\]The tail lies between the two integrals.
\[\cb{\int_{n+1}^{\infty}f(x)\,dx}\le R_n\le\cb{\int_{n}^{\infty}f(x)\,dx}\]Add \(S_n\) to all three parts to bound the sum.
\[\begin{aligned}S_n+\int_{n+1}^{\infty}f\,dx&\le s\\ s&\le S_n+\int_{n}^{\infty}f\,dx\end{aligned}\]This needs \(f\) to be positive, continuous and decreasing for \(x\ge n\), with \(f(k)=a_k\). It also needs the series to converge, so that \(s\) exists.
When it fails
Replace \(\frac1{x^4}\) by a curve like \(\frac1x\). The areas from \(n\) and from \(n+1\) to infinity are both infinite.
The series then diverges, so there is no sum \(s\) to bound. The bounds are useful only when the series converges.
Example 1: A sum between two numbers
Bound the sum of \(\displaystyle\sum_{n=1}^{\infty}ne^{-n}\), using \(S_5\approx0.89486\).
Take \(f(x)=xe^{-x}\), which is positive and continuous. Its derivative is negative for \(x>1\), so \(f\) is decreasing for \(x\ge5\).
\[f'(x)=(1-x)e^{-x}\]Integrate by parts to find the area from \(n\) to infinity.
\[\int_n^{\infty}xe^{-x}\,dx=(n+1)e^{-n}\]For \(n=5\), the upper bound uses the integral from 5 and the lower bound the integral from 6.
\[\begin{aligned}\cb{7e^{-6}}\approx0.0174&\le R_5\\ R_5&\le\cb{6e^{-5}}\approx0.0404\end{aligned}\]Add \(S_5\) to each part.
\[0.9122\le s\le0.9353\]
Example 2: Where the tail starts
Bound the tail \(R_4\) of \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+1)^3}\).
The tail starts with the term \(a_5\). An antiderivative of \(\dfrac{1}{(x+1)^3}\) is \(-\dfrac{1}{2(x+1)^2}\), so the area from \(c\) to infinity is \(\dfrac{1}{2(c+1)^2}\).
\[\int_c^{\infty}\frac{dx}{(x+1)^3}=\frac{1}{2(c+1)^2}\]A tempting wrong move: The tail starts at \(a_5\), so the upper bound is the integral from 5.
\[R_4\le\frac{1}{2\cdot6^2}\approx0.0139\]As a check, the tail is about 0.0164, which is larger than 0.0139. Under the curve, the rectangle for \(a_5\) lies between \(x=4\) and \(x=5\), so the integral from 5 leaves it out.
The rectangle for \(a_5\) starts at \(x=4\), so the upper bound uses the integral from 4. The integral from 5 is the lower bound.
\[\begin{aligned}\frac1{72}\approx0.0139&\le R_4\\ R_4&\le\frac1{50}=0.02\end{aligned}\]
Practice
How many terms of \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n-1)^3}\) guarantee that \(S_n\) is within 0.001 of the sum?
The error is the tail \(R_n\). Use its upper bound, the integral from \(n\), and solve for \(n\).
Solution
The function \(\dfrac{1}{(3x-1)^3}\) is positive, continuous and decreasing for \(x\ge1\), so the shaded area from \(n\) to infinity is an upper bound for the tail.
\[\begin{aligned}R_n&\le\int_n^{\infty}\frac{dx}{(3x-1)^3}\\ &=\frac{1}{6(3n-1)^2}\end{aligned}\]The bound must be less than 0.001. This means \((3n-1)^2>\frac{1000}{6}\approx166.7\), so \(3n-1>12.9\).
\[\frac{1}{6(3n-1)^2}<0.001\]This means \(n>4.6\). For \(n=5\) the bound is \(\frac{1}{6\cdot196}\approx0.00085\). For \(n=4\) it is \(\frac{1}{6\cdot121}\approx0.00138\), which is too large. Five terms guarantee the accuracy.
Questions
The graph shows the curve \(f(x)=\dfrac1{x^2}\) for the series \(\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}\). The rectangles are the terms of the tail \(R_4\), which starts with \(a_5\).
The rectangles lie under the curve and start at \(x=4\). Which integral does the picture give as an upper bound for \(R_4\)?
- \(\displaystyle\int_4^{\infty}\frac{dx}{x^2}\)
- \(\displaystyle\int_5^{\infty}\frac{dx}{x^2}\)
- \(\displaystyle\int_1^{\infty}\frac{dx}{x^2}\)
Answer
\(\displaystyle\int_4^{\infty}\frac{dx}{x^2}\)
The rectangles lie above the curve and start at \(x=5\). What does the picture show?
- \(R_4\ge\displaystyle\int_5^{\infty}\frac{dx}{x^2}\)
- \(R_4\le\displaystyle\int_5^{\infty}\frac{dx}{x^2}\)
- \(R_4=\displaystyle\int_5^{\infty}\frac{dx}{x^2}\)
Answer
\(R_4\ge\displaystyle\int_5^{\infty}\frac{dx}{x^2}\)
Which is the integral-test upper bound for the tail \(R_6\) of \(\displaystyle\sum_{n=1}^{\infty}e^{-2n}\)?
- \(\dfrac{e^{-12}}{2}\)
- \(\dfrac{e^{-14}}{2}\)
- \(\dfrac{e^{-6}}{2}\)
Answer
\(\dfrac{e^{-12}}{2}\)
Summary
You add the integral from \(n+1\) to infinity to \(S_n\) for the lower bound, and the integral from \(n\) to infinity for the upper bound.
This works because the rectangle for the first tail term lies between \(x=n\) and \(x=n+1\) under the curve, and between \(x=n+1\) and \(x=n+2\) above it.
You use it for a convergent series to estimate its sum or to find how many terms give the accuracy you need.
This is one of four skills in topic 10.4 of Which & Why: AP Calculus.