Math Infinitum

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Intuition

A finite series is an ordinary sum written compactly. The summation sign gives a rule for the terms and a range for the index. The index itself is a placeholder: renaming it or shifting it changes the bookkeeping and never the value. One pattern is worth recognising at once, where consecutive terms cancel and almost nothing survives.

The index is the word "guest" in a recipe that says to allow one portion per guest. Renaming it to "visitor" changes no quantity. Starting the count at zero rather than one changes no quantity either, so long as the rule is shifted to match.

Summation notation

The letter below the sign is the index, the numbers below and above give its range, and the expression to the right is the rule. Everything here is a finite sum, so no question of convergence arises yet.

\sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n

Rules that follow at once

The index is a placeholder: $\sum_{k=1}^{n} a_k$ and $\sum_{j=1}^{n} a_j$ are the same number.

A telescoping sum

\sum_{k=1}^{n} (a_{k+1} - a_k) = a_{n+1} - a_1

Write the sum out and watch the interior terms pair off. Each one is added once and subtracted once, so only the two ends are left standing.

Proof steps

\begin{aligned}\sum_{k=1}^{n} (a_{k+1} - a_k) &= (a_2 - a_1) + (a_3 - a_2)\\ &\quad + \dots + (a_{n+1} - a_n)\end{aligned}

Write the sum out in full, one bracket for each value of the index.

\text{Each } a_k \text{ with } 2 \le k \le n \text{ appears twice}

It is added by one bracket and subtracted by the next.

\text{Those pairs cancel}

A number added and subtracted contributes nothing.

\text{Only } -a_1 \text{ and } a_{n+1} \text{ are left}

The first subtraction and the last addition have no partner.

= a_{n+1} - a_1

However long the sum, it collapses to the difference of its ends.

Applications

Practice

The index is a placeholder

\sum_{k=1}^{3} k^2 = \sum_{j=1}^{3} j^2 = 14

Both count the same three squares; only the label differs.

Try it

Which sum equals $\sum_{k=1}^{4} k$ ?

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A shift moves range and rule together

\sum_{k=1}^{3} a_k = \sum_{k=0}^{2} a_{k+1}

Both add $a_1, a_2, a_3$ ; only the labels on the index have moved.

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Which sum equals $\sum_{k=2}^{5} a_k$ ?

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Constants come out

\sum_{k=1}^{n} 3a_k = 3\sum_{k=1}^{n} a_k

Every term carried the same factor of three.

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Which equals $\sum_{k=1}^{n} 5a_k$ ?

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Spot the cancellation

\sum_{k=1}^{n} \big( \tfrac{1}{k} - \tfrac{1}{k+1} \big) = 1 - \tfrac{1}{n+1}

Every interior reciprocal is subtracted as soon as it is added.

Try it

What is $\sum_{k=1}^{5} (a_{k+1} - a_k)$ ?

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What is $\sum_{k=3}^{2} a_k$ ?

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What is $\sum_{k=1}^{4}(2k-1)$ ?

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Sums Split and Release Constants

Both rules come from the ordinary laws of addition and let a complicated sum be broken into pieces.

\sum(a_k+b_k)=\sum a_k+\sum b_k

\sum c\,a_k=c\sum a_k

Try it

Which equals $\sum_{k=1}^{n}(3k+2)$ ?

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Reversing the order of the terms of a finite sum can change its value.

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What You Learned

The index is a placeholder and a shift moves the range and the rule together.
Sums split over addition and release constant factors.
An empty sum is zero.
A telescoping sum collapses to its two ends, and the order of a finite sum is irrelevant.

Final checkpoint

Try it

What is $\sum_{k=1}^{n}\Big(\tfrac{1}{k}-\tfrac{1}{k+1}\Big)$ ?

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Completion

Lesson complete

Great work! You now know how to:

Read and shift summation notation
Split a sum and pull constants out of it
Collapse a telescoping sum to its two ends