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Summation

Along with every sequence a1,a2,a3,…,an,…a_1, a_2, a_3, \ldots, a_n, \ldots we can consider the sequence (Sn)(S_n):

S1=a1,S2=a1+a2,S3=a1+a2+a3,…,Sn=a1+a2+…+an,…S_1 = a_1, \quad S_2 = a_1 + a_2, \quad S_3 = a_1 + a_2 + a_3, \quad \ldots, \quad S_n = a_1 + a_2 + \ldots + a_n, \quad \ldots

Finding the formula for the nn-th term of (Sn)(S_n) is called summation of the first nn terms of the sequence (an)(a_n).

Sigma Notation

📐Definition — Sigma Notation

Using the Greek letter Σ\Sigma (sigma), the sum a1+a2+…+ana_1 + a_2 + \ldots + a_n is written as:

∑k=1nak=a1+a2+…+an.\sum_{k=1}^{n} a_k = a_1 + a_2 + \ldots + a_n.

For example:

12+22+32+…+n2=∑k=1nk2,13+23+33+…+n3=∑k=1nk3.1^2 + 2^2 + 3^2 + \ldots + n^2 = \sum_{k=1}^{n} k^2, \qquad 1^3 + 2^3 + 3^3 + \ldots + n^3 = \sum_{k=1}^{n} k^3.

Properties of Summation

⚡Theorem — Properties of Sigma Notation

  1. ∑k=1nc=nc\displaystyle\sum_{k=1}^{n} c = nc (sum of nn identical terms)

  2. ∑k=1n(ak+bk)=∑k=1nak+∑k=1nbk\displaystyle\sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k (sum of sums equals sum of respective sums)

  3. ∑k=1nc⋅ak=c⋅∑k=1nak\displaystyle\sum_{k=1}^{n} c \cdot a_k = c \cdot \sum_{k=1}^{n} a_k (a constant factor can be taken outside the summation sign)

Standard Sums

⚡Theorem — Standard Sum Formulas
∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}∑k=1nk3=(n(n+1)2)2\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2

These formulas allow us to sum the first nn terms of the sequences of natural numbers, their squares, and their cubes respectively.

Telescoping Sums

ℹNote — Telescoping Sums

If for a given sequence (an)(a_n) one can find a sequence (bn)(b_n) such that an=bn+1−bna_n = b_{n+1} - b_n, then the sum ∑k=1nak\displaystyle\sum_{k=1}^{n} a_k is easy to compute:

a1+a2+…+an=(b2−b1)+(b3−b2)+…+(bn+1−bn)=bn+1−b1.a_1 + a_2 + \ldots + a_n = (b_2 - b_1) + (b_3 - b_2) + \ldots + (b_{n+1} - b_n) = b_{n+1} - b_1.

Examples

✎Example — Sum Using Standard Formulas

Problem. Find the sum 32+94+258+…+2n⋅n+12n\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{25}{8} + \ldots + \dfrac{2^n \cdot n + 1}{2^n}.

Solution. We have: 2k⋅k+12k=k+12k\dfrac{2^k \cdot k + 1}{2^k} = k + \dfrac{1}{2^k}.

Then the given sum can be rewritten as:

(1+12)+(2+122)+(3+123)+…+(n+12n)\left(1 + \frac{1}{2}\right) + \left(2 + \frac{1}{2^2}\right) + \left(3 + \frac{1}{2^3}\right) + \ldots + \left(n + \frac{1}{2^n}\right)=(1+2+3+…+n)+(12+122+123+…+12n)=n(n+1)2+1−12n.= (1 + 2 + 3 + \ldots + n) + \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^n}\right) = \frac{n(n+1)}{2} + 1 - \frac{1}{2^n}.

Therefore, ∑k=1n2k⋅k+12k=n(n+1)2+1−12n\displaystyle\sum_{k=1}^{n} \frac{2^k \cdot k + 1}{2^k} = \frac{n(n+1)}{2} + 1 - \frac{1}{2^n}.

✎Example — Sum 1 + 12 + 45 + ... Using Cubes

Problem. Find the sum 1+12+45+…+n2(2n−1)1 + 12 + 45 + \ldots + n^2(2n - 1).

Solution. Write: k2(2k−1)=2k3−k2k^2(2k - 1) = 2k^3 - k^2. Then

1+12+45+…+n2(2n−1)=∑k=1n(2k3−k2)=2∑k=1nk3−∑k=1nk2=1 + 12 + 45 + \ldots + n^2(2n-1) = \sum_{k=1}^{n}(2k^3 - k^2) = 2\sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} k^2 ==2⋅(n(n+1)2)2−n(n+1)(2n+1)6=n(n+1)(3n2+n−1)6.= 2 \cdot \left(\frac{n(n+1)}{2}\right)^2 - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(3n^2 + n - 1)}{6}.

Answer: n(n+1)(3n2+n−1)6\dfrac{n(n+1)(3n^2 + n - 1)}{6}.

✎Example — Telescoping Sum of Fractions

Problem. Prove that if (an)(a_n) is an arithmetic progression with nonzero terms, then

1a1a2+1a2a3+…+1anan+1=na1an+1.\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_n a_{n+1}} = \frac{n}{a_1 a_{n+1}}.

Solution. We have: 1anan+1=(1an−1an+1)⋅1d\dfrac{1}{a_n a_{n+1}} = \left(\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}\right) \cdot \dfrac{1}{d}, where dd is the common difference. Then

1a1a2+1a2a3+…+1anan+1=1d(1a1−1a2)+1d(1a2−1a3)+…+1d(1an−1an+1)=\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_n a_{n+1}} = \frac{1}{d}\left(\frac{1}{a_1} - \frac{1}{a_2}\right) + \frac{1}{d}\left(\frac{1}{a_2} - \frac{1}{a_3}\right) + \ldots + \frac{1}{d}\left(\frac{1}{a_n} - \frac{1}{a_{n+1}}\right) ==1d(1a1−1an+1)=an+1−a1d⋅a1⋅an+1=dnd⋅a1⋅an+1=na1an+1.= \frac{1}{d}\left(\frac{1}{a_1} - \frac{1}{a_{n+1}}\right) = \frac{a_{n+1} - a_1}{d \cdot a_1 \cdot a_{n+1}} = \frac{dn}{d \cdot a_1 \cdot a_{n+1}} = \frac{n}{a_1 a_{n+1}}.
✎Example — Sum of Factorials

Problem. Find the sum 1⋅1!+2⋅2!+3⋅3!+…+n⋅n!1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n!.

Solution. Note that n⋅n!=(n+1)!−n!n \cdot n! = (n+1)! - n!. Then:

∑k=1nk⋅k!=(2!−1!)+(3!−2!)+(4!−3!)+…+((n+1)!−n!)=(n+1)!−1.\sum_{k=1}^{n} k \cdot k! = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + ((n+1)! - n!) = (n+1)! - 1.

Answer: (n+1)!−1(n+1)! - 1.

Exercises

✏Exercise — Summation Problems

36.1. Find the sum:

  1. 2+10+30+…+n(n2+1)2 + 10 + 30 + \ldots + n(n^2 + 1)
  2. 1⋅2⋅3+2⋅3⋅4+…+n(n+1)(n+2)1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + n(n+1)(n+2)

36.2. Find the sum 1⋅2+2⋅3+…+n(n+1)1 \cdot 2 + 2 \cdot 3 + \ldots + n(n+1).

36.7. Find the sum:

  1. 11⋅5+15⋅9+19⋅13+…+1(4n−7)(4n−3)\dfrac{1}{1 \cdot 5} + \dfrac{1}{5 \cdot 9} + \dfrac{1}{9 \cdot 13} + \ldots + \dfrac{1}{(4n-7)(4n-3)}
  2. 31⋅2+72⋅3+133⋅4+…+n2+n+1n(n+1)\dfrac{3}{1 \cdot 2} + \dfrac{7}{2 \cdot 3} + \dfrac{13}{3 \cdot 4} + \ldots + \dfrac{n^2 + n + 1}{n(n+1)}