Sequences and Series: Chapter Review

Vocabulary and notation

Sequence

An ordered list of numbers, each one a term. Order matters: changing the order can give a different sequence, as reversing 2, 5, 8 does.

Position and the subscript $a_n$

$a_n$ is the term in position $n$. The subscript counts position, never a multiplication; $n$ is always a counting number.

Series, partial sum $S_n$

A series is the sum of a sequence's terms, a single number rather than a list. $S_n$, the $n$th partial sum, sums just the first $n$ terms.

Arithmetic sequence, common difference $d$

Each term is the one before it plus a fixed $d$, which may be positive, negative, fractional, or zero (a constant sequence).

Geometric sequence, common ratio $r$

Each term is the one before it times a fixed $r$. Every $r$ but $0$ is allowed, and with $a_1 \neq 0$ no term is ever $0$.

Explicit form and recursive form

Explicit (closed) form gives a term straight from its position; recursive form builds each term from the one before it.

Summation (sigma) notation

In $\sum_{k=1}^{n} a_k$, $k$ is the index, $1$ and $n$ the lower and upper limits, $a_k$ the summand. The index is a dummy variable: the letter never changes the value.

Converge, diverge

An infinite series converges when its partial sums $S_n$ approach one fixed value as $n$ grows, which becomes the infinite sum; otherwise it diverges and has no finite sum.

Telescoping sum

A sum whose terms are differences of shifted values of one sequence $b_k$ (consecutive values are the gap-1 case), so the interior cancels and a fixed number of boundary values survive.

Formulas and theorems

• Testing arithmetic against geometric

  Arithmetic when every $a_{k+1} - a_k$ gives the same $d$; geometric when every $\frac{a_{k+1}}{a_k}$ gives the same $r$.

  Use when One mismatched pair disqualifies it. Subtract or divide a term by the one BEFORE it; the ratio test needs nonzero terms.

  e.g. $2, 4, 8, 16$: differences $2, 4, 8$ vary, but every ratio is $2$, so geometric.

• $n$th term of an arithmetic sequence

  $$a_n = a_1 + (n-1)d$$

  Use when Any arithmetic sequence, any $n \ge 1$. The step count is $n - 1$, not $n$, because the first term takes no step; at $n = 1$ it returns $a_1$.

• $n$th term of a geometric sequence

  $$a_n = a_1 r^{\,n-1}$$

  Use when Any geometric sequence, any $n \ge 1$, with $r \neq 0$. The exponent is $n - 1$, not $n$, and $r^0 = 1$ returns $a_1$.

  e.g. $a_1 = 7$, $r = 3$: $a_4 = 7 \cdot 3^3 = 189$.

• What the common ratio does

  $r > 1$: the terms grow in size without bound. $0 < r < 1$: they shrink toward $0$, never reaching it. $r < 0$: they alternate in sign. $r = 1$: every term equals $a_1$.

  Use when Any geometric sequence with $a_1 \neq 0$; $r = 0$ is excluded outright. Size follows the powers of $|r|$, so a negative $r$ with $|r| > 1$ alternates AND grows.

• Moving between any two terms

  $$\begin{gathered} a_k = a_m + (k-m)d \\ a_k = a_m\,r^{\,k-m} \end{gathered}$$

  Use when Any positions $m$ and $k$, arithmetic above and geometric below; a negative $k - m$ steps backward. An EVEN gap leaves an even power of $r$, hiding a sign: $r^2 = 4$ allows both $2$ and $-2$.

  e.g. $a_3 = 20$ and $a_7 = 44$: $44 = 20 + 4d$, so $d = 6$.

• Recursive rules

  $$a_n = a_{n-1} + d, \qquad a_n = r\,a_{n-1}$$

  Use when Both for $n \ge 2$, each needing a stated $a_1$.

• Expanding summation notation

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

  Use when The index steps by one from the lower limit through the upper, both included.

  e.g. $\sum_{k=1}^{4} (2k + 1) = 3 + 5 + 7 + 9 = 24$.

• Finite arithmetic series

  $$\begin{gathered} S_n = \frac{n(a_1 + a_n)}{2} \\ = \frac{n}{2}\big(2a_1 + (n-1)d\big) \end{gathered}$$

  Use when $n$ a positive whole number of arithmetic terms; the two forms give the same total.

  e.g. $a_1 = 1$, $a_n = n$ gives $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$, so $1 + 2 + \cdots + 100 = 5050$.

• Counting and summing a range of terms

  The terms from position $m$ to position $n$ number $n - m + 1$, and their total is $S_n - S_{m-1}$.

  Use when $1 \le m \le n$, reading $S_0 = 0$ when $m = 1$. Both endpoints count, hence the $+1$.

  e.g. Rows $7$ through $19$ number $19 - 7 + 1 = 13$.

• Finite geometric series

  $$S_n = \frac{a_1(1 - r^{\,n})}{1 - r} = \frac{a_1(r^{\,n} - 1)}{r - 1}$$

  Use when $r \neq 1$: there the denominator is zero, and since every term equals $a_1$ the sum is simply $n\,a_1$. The two forms agree everywhere, the first tidier when $r < 1$, the second when $r > 1$.

• Infinite geometric series

  $$S = \frac{a_1}{1 - r}$$

  Use when $|r| < 1$ ONLY, a condition on size, so it covers negative ratios like $r = -\tfrac{1}{2}$. When $|r| \ge 1$ there is no finite sum: $r = 1$ runs off, $r = -1$ flips between $a_1$ and $0$ forever, and $|r| > 1$ grows in size past every bound.

• Telescoping collapse

  $$\begin{gathered} \sum_{k=1}^{n} (b_k - b_{k+1}) = b_1 - b_{n+1} \\ \sum_{k=1}^{n} (b_{k+1} - b_k) = b_{n+1} - b_1 \end{gathered}$$

  Text description

  The four brackets of a telescoping sum are stacked in a staircase so that each subtracted term sits directly above the identical added term in the row below; every one of those interior pairs is struck through, leaving only the first term and the last subtracted term, whose difference is the whole sum.

  Use when Every term must be a difference of SHIFTED values of one $b_k$: consecutive values ($b_k - b_{k+1}$) are the gap-1 case, and a wider gap leaves more survivors per end. In the gap-1 case the back survivor is $b_{n+1}$, not $b_n$. The infinite version is finite exactly when the complete surviving boundary expression settles at one fixed value, not necessarily zero.

• The two standard fraction telescopes

  $$\begin{gathered} \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \\ \frac{1}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right) \end{gathered}$$

  Use when Verify a split by recombining over a common denominator; the gap-$2$ split needs the $\tfrac{1}{2}$, since $\frac{1}{k} - \frac{1}{k+2} = \frac{2}{k(k+2)}$. Collapsed they total $1 - \frac{1}{n+1} = \frac{n}{n+1}$ (infinite value $1$) and $\frac{3}{4} - \frac{1}{2}\left(\frac{1}{n+1} + \frac{1}{n+2}\right)$ (two survivors each end; infinite value $\tfrac{3}{4}$).

• Square-root and consecutive-square telescopes

  $$\begin{gathered} \sum_{k=1}^{n} \left(\sqrt{k+1} - \sqrt{k}\right) = \sqrt{n+1} - 1 \\ \sum_{k=1}^{n} (2k+1) = (n+1)^2 - 1 \end{gathered}$$

  Use when Mirror-direction telescopes, $b_k = \sqrt{k}$ and $b_k = k^2$, the second from $(k+1)^2 - k^2 = 2k+1$. Rationalizing turns $\frac{1}{\sqrt{k} + \sqrt{k+1}}$ into $\sqrt{k+1} - \sqrt{k}$, and $\log\!\left(\frac{k+1}{k}\right)$ telescopes to $\log(n+1)$. The square-root sum has NO finite infinite value: $\sqrt{n+1}$ runs off.

Problem types, step by step

Exam traps

• Trap Reporting a term when the question asked for a sum, or the reverse: for $2, 5, 8, 11$, answering "the sum of the first four terms" with $11$.

  Fix $11$ is $a_4$, the fourth TERM; the series is $2 + 5 + 8 + 11 = 26$. Check for the word term or sum before picking a formula.

• Trap Off-by-one on an index: writing $a_n = a_1 + nd$, counting the $10$th through the $30$th terms as $30 - 10 = 20$ terms, or stopping a gap-$1$ telescope at $b_n$.

  Fix Each is one item out. $a_1 + nd$ is really $a_{n+1}$, since the first term takes no step; the range holds $30 - 10 + 1 = 21$ terms because both ends count; a gap-$1$ telescope's last survivor is $b_{n+1}$, not $b_n$ (a wider gap leaves more back survivors still).

• Trap Reaching for $\frac{n(a_1 + a_n)}{2}$ to total a geometric series.

  Fix That averages the first and last terms, exact only for evenly spaced ones. Geometric terms are scaled, not stepped, so use $\frac{a_1(1 - r^{\,n})}{1 - r}$.

• Trap Hunting for $r$ by subtracting consecutive terms, or by dividing a term by the one AFTER it.

  Fix The ratio is a term over the one before it, $r = \frac{a_2}{a_1}$; the other way gives the reciprocal. In $3, 6, 12, 24$ every ratio is $2$ while the differences never settle.

• Trap Dropping the minus sign of a negative ratio inside $\frac{a_1}{1 - r}$.

  Fix With $a_1 = 9$ and $r = -\tfrac{1}{3}$ the denominator is $1 - \left(-\tfrac{1}{3}\right) = \tfrac{4}{3}$, so the sum is $\tfrac{27}{4}$, not the $\tfrac{27}{2}$ that $1 - \tfrac{1}{3}$ would give.

• Trap Dividing an inequality by $\log r$ without flipping it: reading $300 \cdot 0.6^{\,n-1} < 20$ as $n - 1 < \frac{\log(1/15)}{\log 0.6}$, then rounding down.

  Fix For $0 < r < 1$ that logarithm is negative, so dividing reverses the sign: $n - 1 > 5.30$, and a position rounds UP, giving $n = 7$. The $6$th rebound is $23.3$ cm, still above $20$.

• Trap Totaling a bouncing ball's travel as one series: dropped from $12$ m and rebounding to half, answering $\frac{12}{1 - \tfrac{1}{2}} = 24$ m.

  Fix The drop is traveled once, every rebound twice, up and down: $12 + 2\left(\frac{6}{1 - \tfrac{1}{2}}\right) = 12 + 24 = 36$ m.

• Trap Dropping a number into the wrong slot: reading "the $12$th term is $40$" as $n = 40$, or putting the term count where $r$ belongs.

  Fix $n$ is only ever a position or a count, $r$ is the multiplier, $a_n$ and $S_n$ are values. Here $n = 12$ goes in and $40$ is what the formula must produce.

• Trap Leaving one survivor at each end of a gap-$2$ telescope, so $\sum_{k=1}^{n} \frac{1}{k(k+2)}$ is read as $\frac{1}{2}\left(1 - \frac{1}{n+2}\right)$.

  Fix With a gap of $2$ each piece cancels two brackets later, so inside the $\tfrac{1}{2}$ bracket the survivors are $1$ and $\tfrac{1}{2}$ in front, $\frac{1}{n+1}$ and $\frac{1}{n+2}$ at the back. At $n = 2$ the correct $\tfrac{11}{24}$ beats the wrong $\tfrac{3}{8}$.