Summation of series
Introduced| English | Meaning |
|---|---|
| summation/sʌˈmeɪʃn/ | Adding the terms of a sequence over a stated range of indices. |
| method of differences/ˈmeθəd ɒv ˈdɪfrənsɪz/ | In summing series, rewriting terms as differences so that most terms cancel and only boundary terms remain. |
| telescope/ˈtelɪskəʊp/ | To cancel matching terms of opposite signs in a sum, leaving its boundary terms. |
| partial fractions/ˈpɑːʃl ˈfrækʃnz/ | A way to rewrite a rational expression as simpler fractions, often making its integral easier. |
| converge/kənˈvɜːdʒ/ | A sequence converges when its terms approach a finite limit; a series converges when its partial sums do. |
| sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ | The limiting value of the partial sums of a convergent infinite series. |
Count work without adding every stage
- An algorithm performs $r^2$ operations at stage $r$. To estimate its total work after thousands of stages, we need one expression for the whole sum.
- Summation 求和 can combine standard formulas or cancel shifted terms. An infinite total needs a separate check: a finite formula alone does not prove convergence.
Standard sums and their bounds
- For a positive integer $n$, the standard results are:
- The limits matter: $\sum_{r=1}^n1=n$, but $\sum_{r=0}^n1=n+1$. Check the number of terms before applying a formula. The cube identity uses consecutive integers from 1; it is not true for arbitrary lists of numbers.
Add consecutive integers
Sₙ = 1 + 2 + ⋯ + n
With a=1 and d=1, the terms are consecutive integers and Sₙ=n(n+1)/2. Their partial sums grow without bound; a finite sum is not automatically a convergent infinite series.
Using Σr = ½n(n+1), what is the sum of the first 5 integers?
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½ × 5 × 6 = 15 (= 1+2+3+4+5).
Using Σr² = (1/6)n(n+1)(2n+1), what is the sum of the first 3 squares?
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(1/6)(3)(4)(7) = 84/6 = 14 (= 1+4+9).
Using Σr³ = ¼n²(n+1)², what is the sum of the first 3 cubes?
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¼ × 9 × 16 = 36 (= 1+8+27). Note: 36 = 6² = (1+2+3)².
Complete the standard result: Σr (from 1 to n) = ______ n(n+1).
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The sum of the first n integers is ½n(n+1).
Expand first, then sum each part
- For $\sum_{r=1}^n(2r+1)(r+2)$, expand to $2r^2+5r+2$. The result is $2\sum r^2+5\sum r+2n$.
- Substitute the standard results and simplify:
At $n=2$, the original terms are $3\times3$ and $5\times4$, totalling 29. The formula also gives $2(16+42+29)/6=29$. A small-$n$ check catches missing constants and index errors.
Σ(2r+1) from r=1 to n = n(n+2). For n=4, find the sum.
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4 × 6 = 24. Check: 3+5+7+9 = 24.
Find $\sum_{r=1}^n(3r-2)(2r+1)$ as a polynomial in n and verify your expression at n=2.
Expand to 6r²−r−2. Using standard sums gives $n(n+1)(2n+1)-n(n+1)/2-2n=n(4n^2+5n-3)/2$. At n=2 this is23; the original terms are3 and20, also23.
Change the starting or ending index
- If $F(m)=\sum_{r=1}^m u_r$, then $\sum_{r=p}^q u_r=F(q)-F(p-1)$. The last removed term is $p-1$, not $p$.
- For $\sum_{r=0}^n(r+1)^2$, put $k=r+1$. The new bounds are $1$ and $n+1$, so the sum is $(n+1)(n+2)(2n+3)/6$. Change the limits as well as the expression.
Find $\sum_{r=n+1}^{2n}(r^2+2r)$ in terms of the positive integer n. Check n=1.
Let $F(m)=m(m+1)(2m+7)/6$. The required sum is F(2n)−F(n), giving $n(14n^2+27n+7)/6$. At n=1 only r=2 occurs, giving8, which agrees.
Find $\sum_{r=0}^n(r+1)^3$ by changing the index, and confirm it by expanding the summand. State how many terms there are.
With k=r+1 the limits are1 to n+1, so the sum is $[(n+1)(n+2)/2]^2$. Expansion gives $\sum r^3+3\sum r^2+3\sum r+\sum1$ over0..n, whose last contribution is n+1. These simplify to the same result. There are n+1 terms.
The method of differences 差分法
- Write a term as $F(r)-F(r+1)$, then show the first few terms and the last one. The middle terms telescope 裂项相消, leaving the boundary values $F(1)-F(n+1)$.
- A wider shift leaves more boundary terms. For $F(r)-F(r+2)$, the two starting terms and two ending terms survive:
Partial fractions 部分分式 make cancellation visible
- For $1/[r(r+2)]$, solve $1=A(r+2)+Br$: $A=1/2,B=-1/2$. Thus each term is $\tfrac12(1/r-1/(r+2))$.
- Use the two-step cancellation, keeping its factor:
A decomposition is not yet a sum. Show enough first and last terms to establish the cancellation, especially when three partial fractions or an index shift are involved. Check the formula at $n=1$: it gives $1/3$, the original first term.
Use partial fractions and differences to find $\sum_{r=1}^n1/[(3r-1)(3r+2)]$ and its sum to infinity.
The term is $\tfrac13[1/(3r-1)-1/(3r+2)]$. Write the first two and last terms; all intermediate denominators cancel. The sum is $\tfrac13[1/2-1/(3n+2)]=n/[2(3n+2)]$. Its limit is1/6.
Find $\sum_{r=2}^n1/[r(r+2)]$ for n≥2 and its sum to infinity. Show all surviving boundary terms.
The term is $\tfrac12(1/r-1/(r+2))$. The sum is $\tfrac12[1/2+1/3-1/(n+1)-1/(n+2)]=5/12-1/[2(n+1)]-1/[2(n+2)]$. Its limit is5/12.
Decompose $(2r+1)/[r(r+1)(r+2)]$ and use differences to find its sum from r=1 to n and to infinity.
The decomposition is $1/(2r)+1/(r+1)-3/[2(r+2)]$. On adding the shifted lists the interior coefficients cancel. The surviving total is $5/4-1/[2(n+1)]-3/[2(n+2)]$. At n=1 this is1/2, matching the original first term. Its limit is5/4.
Take the limit of the finite sum
- A series converges 收敛 when its partial sums approach a finite real limit. In the last example the two ending fractions tend to zero, so its sum to infinity 无穷和 is $3/4$.
- The remaining tail after $n$ terms is $S_\infty-S_n$. Here it is $\dfrac1{2(n+1)}+\dfrac1{2(n+2)}$, which can be used to choose how many terms give a required accuracy.
Terms tending to zero is not enough. For $u_r=\sqrt{r+1}-\sqrt r$, the terms tend to zero, but $S_n=\sqrt{n+1}-1$ grows without bound. A finite partial-sum limit is the deciding evidence.
In the method of differences, if each term is f(r) − f(r+1), most terms cancel (telescope).
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Consecutive terms cancel, leaving only the first and last pieces.
The finite sum of 1/(r(r+1)) from r=1 to n is 1−1/(n+1). What is its limit as n→∞?
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As n→∞, 1/(n+1)→0, so the sum approaches 1.
For $S_n=\sum_{r=1}^n1/[r(r+2)]$, find the least positive integer n for which the difference from the infinite sum is less than0.01. Show why it is the least.
The tail is $1/[2(n+1)]+1/[2(n+2)]$, positive and strictly decreasing. At n=98 it is199/19800>1/100; at n=99 it is201/20200<1/100. Hence the least n is99. The next individual term alone would not measure the whole remaining sum.
Differences need not be partial fractions
- Let $P_r=r(r+1)(r+2)$. Factor the common part to obtain $P_{r+1}-P_r=3(r+1)(r+2)$. Hence a sum of $(r+1)(r+2)$ is $(P_{n+1}-P_1)/3$.
- Another route is to put two fractions over one denominator. For example, $1/r^2-1/(r+1)^2=(2r+1)/[r^2(r+1)^2]$. Recognising the difference avoids trying to use a formula for $\sum1/r^2$.
Show that $(3r^2+3r+1)/[r^3(r+1)^3]=1/r^3-1/(r+1)^3$, then sum from r=3 to n and to infinity.
The combined numerator on the right is (r+1)³−r³=3r²+3r+1. The finite sum is $1/27-1/(n+1)^3$ for n≥3, and its limit is1/27. This does not use a general formula for the sum of reciprocal cubes.
Let $P_r=r(r+1)(r+2)(r+3)$. Factor $P_{r+1}-P_r$, then find $\sum_{r=1}^n(r+1)(r+2)(r+3)$. Does the corresponding infinite sum converge?
The difference is $(r+1)(r+2)(r+3)[(r+4)-r]=4(r+1)(r+2)(r+3)$. The sum is $(P_{n+1}-P_1)/4=[(n+1)(n+2)(n+3)(n+4)-24]/4$. It grows without bound, so the infinite series does not converge.
Check parameter endpoints separately
- Let $u_r=t^{r(r+1)}-t^{(r+1)(r+2)}$. The finite sum is $S_n=t^2-t^{(n+1)(n+2)}$. For $|t|<1$, its limit is $t^2$.
- At $t=1$ or $t=-1$, every exponent is even and every term is zero: the infinite sum is 0, not 1. For $|t|>1$, the ending power grows without bound. The convergence set is $[-1,1]$, but the sum needs separate endpoint values.
For real t let $v_r=t^{r^2}-t^{(r+1)^2}$. Find the finite sum from r=1 to n, the full set of t for convergence, and the infinite sum in each convergent case.
The finite sum is $t-t^{(n+1)^2}$. For −1<t<1 the ending power tends to0 and the sum is t. At t=1 every term is0, so the sum is0. At t=−1 the ending powers alternate, and partial sums alternate between−2 and0, so there is no limit. For |t|>1 the endpoint magnitude grows without bound. The convergence set is (−1,1], with a separate value0 at1.
Prove the sum before taking infinity
- Choose standard sums after expanding, or differences after rearranging the term. State the index limits, show the surviving boundaries and check a small value of $n$.
- For an infinite series, inspect the limit of the exact partial sum. Check special parameter values in the original terms; use the remaining tail, not just the next term, for an error bound.
Write the finite partial sum and its uncancelled boundary terms before taking a limit. Check excluded parameter values separately.
What is the sum of the first 4 cubes, and how does it relate to the sum of the first 4 integers?
The cube-sum identity says the sum of the first n cubes is the square of n times (n plus 1) divided by 2.
What is the sum of the first 4 cubes?
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The integers sum to 10, and 10 squared is 100. Other options — 36 (the lesson sum of the first 3 cubes); 10 (the sum of the first 4 integers).
The terms $u_r=\sqrt{r+2}-\sqrt{r+1}$ tend to zero. Determine the exact partial sum and decide whether the series converges, explaining your conclusion.
The partial sum telescopes to $\sqrt{n+2}-\sqrt2$, which grows without bound. Also $u_r=1/(\sqrt{r+2}+\sqrt{r+1})$ tends to0. These facts are compatible: terms tending to0 is necessary but does not guarantee that their accumulated sum has a finite limit.