Distinguish a sequence from a series
A tank filled in stages
A tank takes \(\frac23\) liter of water on day 1, and each later day it takes one third of the amount from the day before. The bars on the left show the pours, a list in a fixed order: \(\frac23\), \(\frac29\), \(\frac2{27}\) and so on.
The column on the right stacks the pours, so its height is the water in the tank. On day 3, the tank holds \(\frac23+\frac29+\frac2{27}=\frac{26}{27}\) liter.
Move the slider. The amounts in the tank after each day form a second list: \(\frac23\), \(\frac89\), \(\frac{26}{27}\), \(\frac{80}{81}\).
The pours become smaller and smaller. The water in the tank rises, but it stays below 1 liter, the height of the dashed line.
Terms, running totals and the total
The bars show any list of numbers \(a_1, a_2, a_3,\dots\). A list in a fixed order is a sequence, and each number in it is a term.
The column stacks the first \(n\) bars, so its height is their total. This total is the partial sum \(S_n\).
Move the slider from \(n\) to \(n+1\). The column gains one piece, the bar \(a_{n+1}\), so each partial sum is the one before it plus one new term.
The partial sums form a second sequence. A series adds all the terms, and its sum is the number that the partial sums approach as \(n\) grows.
Notation
The term \(a_n\) is the number in position \(n\). A different order gives a different sequence.
\[\cy{a_1},\ \cy{a_2},\ \cy{a_3},\ \dots,\ \cy{a_n},\ \dots\]The partial sum \(S_n\) adds the first \(n\) terms. It is one number for each \(n\), so the partial sums also form a sequence.
\[\cb{S_n}=\cy{a_1}+\cy{a_2}+\cdots+\cy{a_n}\]Each partial sum is the one before it plus one term, because the column gains one piece.
\[\cb{S_n}=\cb{S_{n-1}}+\cy{a_n}\]This holds for \(n\ge2\). The first partial sum is \(S_1=a_1\).
The sum \(S\) of the series is the limit of the partial sums, if this limit is a number.
\[\sum_{n=1}^{\infty}\cy{a_n}=\lim_{n\to\infty}\cb{S_n}\]The terms \(a_n\) and the partial sums \(S_n\) are two sequences, but the sum \(S\) of the series is one number.
When it fails
A series has a sum only when its partial sums approach a number. Make every term 1, and the partial sums become 1, 2, 3, and so on.
The column gains 1 with each term and rises past the top of the figure, as the red arrow shows. It approaches no height, so the series has no sum, but the terms still form a sequence.
Example 1: Terms and partial sums
A sequence has the terms \(a_n=3n+1\). Find \(a_5\) and \(S_4\). Then find \(S_5\).
The term \(a_5\) is the number in position 5, so substitute \(n=5\) into the formula.
\[\cy{a_5}=3(5)+1=16\]The partial sum \(S_4\) adds the first four terms, 4, 7, 10 and 13.
\[\cb{S_4}=4+7+10+13=34\]The column gains bar 5, so \(S_5\) is \(S_4\) plus the new term.
\[\cb{S_5}=\cb{S_4}+\cy{a_5}=34+16=50\]
Example 2: The terms and the sum are different numbers
A series has the terms \(a_n=\dfrac{5}{10^n}\). Find its sum.
The first terms are 0.5, 0.05 and 0.005, so the first partial sums are 0.5, 0.55 and 0.555.
A tempting wrong move: The terms approach 0, so the series has the sum 0.
\[\lim_{n\to\infty}a_n=0\ \Rightarrow\ \sum_{n=1}^{\infty}a_n=0\]The red ring marks the claimed sum 0 at the foot of the column. The column is already 0.5 tall after one bar, and every later bar adds a positive height, so the partial sums cannot approach 0.
The sum is the limit of the partial sums, not of the terms. The partial sums 0.5, 0.55, 0.555, and so on approach \(0.555\ldots\), which is \(\frac59\).
\[\sum_{n=1}^{\infty}\frac{5}{10^n}=\frac59\]
Practice
A series has the terms \(a_n=\frac{4}{5^n}\). Find \(a_3\) and \(S_3\). Then find the number that the partial sums approach.
Move the slider. Read the height of the column for \(n=1\), 2 and 3, and how far each height is below 1.
Solution
Substitute \(n=3\) to find the third term.
\[a_3=\frac{4}{5^3}=\frac{4}{125}\]Each partial sum adds the next term to the one before.
\[S_1=\frac45,\quad S_2=\frac{24}{25},\quad S_3=\frac{124}{125}\]Each partial sum is 1 minus a gap: \(\frac15\), \(\frac1{25}\), \(\frac1{125}\). Each gap is one fifth of the one before, so the partial sums approach 1.
\[\sum_{n=1}^{\infty}\frac{4}{5^n}=1\]
Questions
Questions 1 and 2 use the figure. The bars on the left show the terms \(a_n=\frac{3}{2^n}\), and the column on the right stacks them, so its height is the partial sum.
What is the partial sum \(S_3\)?
- 2.625
- 0.375
- 2.25
Answer
2.625
As \(n\) grows, which number do the partial sums \(S_n\) approach?
- 3
- 0
- 1.5
- No number, because the column keeps rising
Answer
3
A series has the terms \(a_n=n^2\). What is \(S_4\)?
- 30
- 16
- 100
Answer
30
Summary
A term \(a_n\) is one bar, and the partial sum \(S_n\) is the height of the column that stacks the first \(n\) bars.
The sum \(S\) of the series is the number that the partial sums approach, not the limit of the terms.
When a question asks for a term, a partial sum or a sum, first decide which of the three it means.
This is one of five skills in topic 10.1 of Which & Why: AP Calculus.