Unit 8, lesson 67

Reading and Rewriting Finite Sums

How to read sigma notation as a finite row of terms, count terms from the bounds, rename or shift the index without changing the sum, and spot a telescoping sum.

  • Theory part 133
  • Practice part 134
  • Indices, Factorials, and Sums

Adding the first hundred odd numbers by hand is tedious, and writing all hundred of them is worse. Sigma notation packs such a sum into one line. Riemann sums, partial sums of series and power series are all written this way, so you need to read the notation fluently, turn it back into an honest row of terms, and rewrite it without changing its value.

Key terms

Summation notation and summand

Definition Let m≤Mm \le M be integers and let aka_k be a number defined for each integer kk with m≤k≤Mm \le k \le M. Then

∑k=mMak=am+am+1+am+2+⋯+aM.\begin{aligned} \sum_{k=m}^{M} a_k &= a_m + a_{m+1} \\[0.4em] &\quad + a_{m+2} + \cdots + a_M. \end{aligned}

The symbol ∑\sum is a capital sigma and means add up. The expression aka_k that follows it is the summand, the formula for a typical term.

In plain words Start the counter at the bottom value, plug in each integer up to the top value, and add all the results.

Example The summand here is 2k+12k + 1.

∑k=14(2k+1)=3+5+7+9=24.\sum_{k=1}^{4} (2k + 1) = 3 + 5 + 7 + 9 = 24.

Parentheses around the summand matter. Without them, ∑k=142k+1\sum_{k=1}^{4} 2k + 1 is ambiguous. The common convention is that the sigma reaches only over the product 2k2k, which gives (∑k=142k)+1=21\left(\sum_{k=1}^{4} 2k\right) + 1 = 21, but not every reader follows it. Put parentheses around any summand with more than one term.

Index of summation

Definition The letter written under the sigma, here kk, is the index of summation. It takes every integer value from the lower bound to the upper bound, one at a time and in steps of 11. It never takes values between those integers.

In plain words The index is the counter that steps through the terms. Each value it takes produces one term.

Example In the sum below the index runs through 0,1,2,30, 1, 2, 3, so the first term is 20=12^0 = 1, not 00.

∑k=032k=1+2+4+8=15.\sum_{k=0}^{3} 2^k = 1 + 2 + 4 + 8 = 15.

Lower and upper bounds

Definition In ∑k=mMak\sum_{k=m}^{M} a_k the integer mm is the lower bound and MM is the upper bound. Both endpoints are included, so the sum has exactly M−m+1M - m + 1 terms. When m=Mm = M the sum has a single term, ama_m.

In plain words Start at the bottom number, stop at the top number, and count both. The number of terms is top minus bottom plus one.

Example The sum

∑k=37k2=9+16+25+36+49=135\begin{aligned} \sum_{k=3}^{7} k^2 &= 9 + 16 + 25 + 36 + 49 \\[0.4em] &= 135 \end{aligned}

has 7−3+1=57 - 3 + 1 = 5 terms. A quick count before you expand catches the common slip of stopping one term early. Also ∑k=22k2=4\sum_{k=2}^{2} k^2 = 4, a sum of one term.

Finite sum

Definition A sum with integer bounds m≤Mm \le M is a finite sum, an ordinary sum of M−m+1M - m + 1 numbers. Because it is ordinary addition, it can be regrouped term by term. For any constant cc and summands aka_k and bkb_k,

∑k=mM(c ak+bk)=c∑k=mMak+∑k=mMbk,\sum_{k=m}^{M} (c\,a_k + b_k) = c \sum_{k=m}^{M} a_k + \sum_{k=m}^{M} b_k,

and a constant summand gives ∑k=mMc=(M−m+1) c\sum_{k=m}^{M} c = (M - m + 1)\,c. Finite sums always have a value, while infinite sums in calculus need limits.

In plain words Pull constant factors out, split sums of terms apart, and remember that adding the same number repeatedly is multiplication by the number of terms.

Example Using the known sum 1+2+3+4+5=151 + 2 + 3 + 4 + 5 = 15,

∑k=15(3k−2)=3∑k=15k−∑k=152=3⋅15−5⋅2=35.\begin{aligned} \sum_{k=1}^{5} (3k - 2) &= 3 \sum_{k=1}^{5} k - \sum_{k=1}^{5} 2 \\[0.4em] &= 3 \cdot 15 - 5 \cdot 2 = 35. \end{aligned}

Writing the row of terms confirms it, 1+4+7+10+13=351 + 4 + 7 + 10 + 13 = 35.

Dummy index

Definition The index of a sum is a dummy index, also called a bound variable. The value of the sum does not depend on its name, so for integers m≤Mm \le M

∑k=mMak=∑j=mMaj.\sum_{k=m}^{M} a_k = \sum_{j=m}^{M} a_j.

The new name must not already be used for something else in the expression. Letters that appear in the bounds or in the summand and are not the index, such as nn in ∑k=1nk\sum_{k=1}^{n} k, are not dummies, and the sum depends on them.

In plain words The index is a placeholder that disappears once you add up. Renaming it changes how the sum looks, not what it equals.

Example Both sums below expand to 1+12+131 + \frac{1}{2} + \frac{1}{3}.

∑k=131k=∑j=131j=116.\sum_{k=1}^{3} \frac{1}{k} = \sum_{j=1}^{3} \frac{1}{j} = \frac{11}{6}.

By contrast ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n + 1)}{2} for each positive integer nn. The answer contains nn and no kk.

Reindexing

Definition Reindexing a sum means shifting the values of the index. With an integer ss, put j=k−sj = k - s, so that k=j+sk = j + s. Then for integers m≤Mm \le M

∑k=mMak=∑j=m−sM−saj+s.\sum_{k=m}^{M} a_k = \sum_{j=m-s}^{M-s} a_{j+s}.

Both bounds change by the same amount, and every occurrence of kk in the summand is replaced by (j+s)(j + s). Renaming alone does not change a sum, but shifting the values does require changing both bounds.

In plain words If the counter now starts lower, the formula has to make up for it, so the same terms come out in the same order.

Example To make ∑k=38(2k−1)\sum_{k=3}^{8} (2k - 1) start at 11, set j=k−2j = k - 2. When k=3k = 3 the new index is j=1j = 1, and when k=8k = 8 it is j=6j = 6.

∑k=38(2k−1)=∑j=16(2(j+2)−1)=∑j=16(2j+3).\begin{aligned} \sum_{k=3}^{8} (2k - 1) &= \sum_{j=1}^{6} \bigl(2(j + 2) - 1\bigr) \\[0.4em] &= \sum_{j=1}^{6} (2j + 3). \end{aligned}

Both rows are 5+7+9+11+13+15=605 + 7 + 9 + 11 + 13 + 15 = 60, and both have 66 terms. Checking the first and last terms is the fastest test of a reindexing.

Telescoping cancellation

Definition Suppose the summand has the form bk−bk+1b_k - b_{k+1} for numbers bm,bm+1,…,bM+1b_m, b_{m+1}, \ldots, b_{M+1}. Then for integers m≤Mm \le M

∑k=mM(bk−bk+1)=bm−bM+1.\sum_{k=m}^{M} (b_k - b_{k+1}) = b_m - b_{M+1}.

Each −bk+1-b_{k+1} cancels the +bk+1+b_{k+1} at the start of the next term, so only the first piece of the first term and the last piece of the last term survive. The same idea gives ∑k=mM(bk+1−bk)=bM+1−bm\sum_{k=m}^{M} (b_{k+1} - b_k) = b_{M+1} - b_m.

In plain words Write out the row and watch neighbors cancel in pairs. Whatever has no partner at the two ends is the answer.

Example For each integer k≥1k \ge 1 the summand splits as

1k(k+1)=1k−1k+1.\frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}.

So for a fixed positive integer NN

∑k=1N1k(k+1)=(1−12)+(12−13)+⋯+(1N−1N+1)=1−1N+1.\begin{aligned} \sum_{k=1}^{N} \frac{1}{k(k + 1)} &= \left(1 - \frac{1}{2}\right) \\[0.4em] &\quad + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots \\[0.4em] &\quad + \left(\frac{1}{N} - \frac{1}{N + 1}\right) \\[0.4em] &= 1 - \frac{1}{N + 1}. \end{aligned}

With N=4N = 4 this predicts 45\frac{4}{5}, and indeed 12+16+112+120=45\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} = \frac{4}{5}.

Check yourself

  1. Write out ∑k=26(−1)kk\sum_{k=2}^{6} (-1)^k k as a row of terms and find its value.
Show the answer

The index runs through 2,3,4,5,62, 3, 4, 5, 6, which gives 55 terms, 2−3+4−5+6=42 - 3 + 4 - 5 + 6 = 4.

  1. Rewrite ∑k=05(3k+1)\sum_{k=0}^{5} (3k + 1) as a sum whose index starts at 11, and check that both have the same value.
Show the answer

Put j=k+1j = k + 1, so k=j−1k = j - 1 and the bounds become 11 and 66.

∑k=05(3k+1)=∑j=16(3(j−1)+1)=∑j=16(3j−2).\begin{aligned} \sum_{k=0}^{5} (3k + 1) &= \sum_{j=1}^{6} \bigl(3(j - 1) + 1\bigr) \\[0.4em] &= \sum_{j=1}^{6} (3j - 2). \end{aligned}

Both rows are 1+4+7+10+13+16=511 + 4 + 7 + 10 + 13 + 16 = 51.

  1. Find ∑k=115(k+1−k)\sum_{k=1}^{15} \left(\sqrt{k + 1} - \sqrt{k}\right) without adding fifteen terms.
Show the answer

This is telescoping with bk=kb_k = \sqrt{k} in the form bk+1−bkb_{k+1} - b_k. Only 16\sqrt{16} from the last term and −1-\sqrt{1} from the first term survive, so the sum is 4−1=34 - 1 = 3.

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