TBIL-Cal Partial Sums and Series (SQ3)

$(1 - x) (1 + x + x^{2}) .$
$(1 - x) (1 + x + x^{2} + x^{3}) .$
$(1 - x) (1 + x + x^{2} + x^{3} + x^{4}) .$
$(1 - x) (1 + x + x^{2} + \dots + x^{n}),$ where $n$ is any nonnegative integer.

🔗

Activity 8.3.7.

🔗

Suppose

S_{5} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} .

Without actually computing this sum, which of the following is equal to

(1 - \frac{1}{2}) S_{5} ?

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} - \frac{1}{64} .$
$1 - \frac{1}{64} .$
$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - \frac{1}{32} .$

🔗

Activity 8.3.8.

🔗

Recall from Activity 8.3.4 that

A_{100} = 2 + \frac{2}{3} + \frac{2}{3^{2}} + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \dots + \frac{2}{3^{100}} = 2 (1 + \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \dots + \frac{1}{3^{100}}) .

🔗

(a)

🔗

Which of the following is equal to

(1 - \frac{1}{3}) A_{100} ?

$1 - \frac{1}{3^{101}} .$
$1 - \frac{1}{3^{100}} .$
$2 (1 - \frac{1}{3^{101}}) .$
$2 (1 - \frac{1}{3^{100}}) .$

🔗

(b)

🔗

Based on your previous choice, write out an expression for

A_{100} .

🔗

Activity 8.3.9.

🔗

Suppose that

{b_{n}}_{n = 0}^{\infty} = {(- 2)^{n}}_{n = 0}^{\infty} = {1, - 2, 4, - 8, \dots} .

Let

B_{n} = \sum_{i = 0}^{n} b_{i}

be the

n

th partial sum of

{b_{n}} .

🔗

(a)

🔗

Find simple expressions for the following:

$(1 - (- 2)) B_{10} .$
$(1 - (- 2)) B_{30} .$
$(1 - (- 2)) B_{n} .$ Choose from the following:
1. $1 + (- 2)^{n} .$
2. $1 - (- 2)^{n} .$
3. $1 + (- 2)^{n + 1} .$
4. $1 - (- 2)^{n + 1} .$
5. $1 - 2^{n} .$

🔗

(b)

🔗

Based on your previous answers, solve for the following:

$B_{10} .$
$B_{30} .$
$B_{n} .$ Choose from the following:
1. $\frac{1 - (- 2)^{n + 1}}{1 - (- 2)}$
2. $\frac{1 - (- 2)^{n + 1}}{1 - 2}$
3. $\frac{1 - (- 2)^{n + 1}}{1 + (- 2)}$
4. $\frac{1 - (- 2)^{n}}{1 - 2}$
5. $\frac{1 - (- 2)^{n}}{1 - (- 2)}$

🔗

Activity 8.3.10.

🔗

Consider the following sequences:

${a_{n}}_{n = 0}^{\infty} = {{(- \frac{2}{3})}^{n}}_{n = 0}^{\infty} .$
${b_{n}}_{n = 0}^{\infty} = {2 \cdot {(- 1)}^{n}}_{n = 0}^{\infty} .$
${c_{n}}_{n = 0}^{\infty} = {- 3 \cdot {(1.2)}^{n}}_{n = 0}^{\infty} .$

🔗

(a)

🔗

Find the closed form for the

n

th partial sum for the geometric sequence

A_{n} = \sum_{i = 0}^{n} a_{i} = \sum_{i = 0}^{n} {(- \frac{2}{3})}^{n} .

$\frac{3}{5} (1 - {(- \frac{2}{3})}^{n + 1}) .$
$\frac{5}{3} (1 - {(- \frac{2}{3})}^{n + 1}) .$
$\frac{5}{3} (1 + \frac{2}{3} {(\frac{2}{3})}^{n}) .$
$\frac{3}{5} (1 + \frac{2}{3} {(\frac{2}{3})}^{n}) .$
$1 - {(- \frac{2}{3})}^{n + 1} .$

🔗

(b)

🔗

Find the closed form for the

n

th partial sum for the geometric sequence

B_{n} = \sum_{i = 0}^{n} b_{i} = \sum_{i = 0}^{n} 2 \cdot {(- 1)}^{n} .

$2^{n + 1} .$
$1 - (- 1)^{n + 1} .$
$1 + (- 1)^{n} .$
$2 (1 + (- 1)^{n}) .$
$2 (1 - (- 1)^{n + 1}) .$

🔗

(c)

🔗

Find the closed form for the

n

th partial sum for the geometric sequence

C_{n} = \sum_{i = 0}^{n} c_{i} = \sum_{i = 0}^{n} - 3 \cdot {(1.2)}^{n} .

🔗

Activity 8.3.11.

🔗

Given the closed forms you found in Activity 8.3.10, which of the following limits are defined? If defined, what is the limit?

$lim_{n \to \infty} A_{n} .$
$lim_{n \to \infty} B_{n} .$
$lim_{n \to \infty} C_{n} .$

🔗

Definition 8.3.12.

🔗

Given a sequence

{a_{n}}_{n = k}^{\infty},

we define its infinite series (or just series) to be its sequence of its partial sums

{A_{n}}_{n = k}^{\infty} = {\sum_{i = k}^{n} a_{i}}_{n = k}^{\infty} = {a_{k}, a_{k} + a_{k + 1}, a_{k} + a_{k + 1} + a_{k + 2}, \dots}

🔗

and often use the notation

\sum_{i = k}^{\infty} a_{i} = a_{k} + a_{k + 1} + a_{k + 2} + \dots

🔗

to represent it. We will also write

\sum a_{i}

for short when the starting index

n = k

is either known from context or irrelevant.

🔗

When the series (the sequence of partial sums) converges to a limit, we say the series is convergent and this limit is the value of the series, and write:

\sum_{i = k}^{\infty} a_{i} = a_{k} + a_{k + 1} + a_{k + 2} + \dots = lim_{n \to \infty} \sum_{i = k}^{n} a_{i} .

🔗

When the series (the sequence of partial sums) diverges, we say the series is divergent.

🔗

Activity 8.3.13.

🔗

Which of the following series are infinite?

$\sum_{n = 0}^{\infty} 3 (0.8)^{n} .$
$\sum_{n = 0}^{\infty} 2 {(\frac{5}{4})}^{n} .$
$\sum_{n = 0}^{\infty} {(\frac{5}{6})}^{n} .$
$\sum_{n = 0}^{\infty} \frac{1}{2} {(81)}^{n} .$
$\sum_{n = 0}^{\infty} 10 {(- \frac{1}{5})}^{n} .$

🔗

Activity 8.3.14.

🔗

Let

{a_{n}}_{n = 1}^{\infty} = {\frac{1}{n} - \frac{1}{n + 1}} = 1 - \frac{1}{2}, \frac{1}{2} - \frac{1}{3}, \frac{1}{3} - \frac{1}{4}, \dots .

Let

A_{n} = \sum_{i = 1}^{n} a_{i} = \sum_{i = 1}^{n} (\frac{1}{i} - \frac{1}{i + 1}) .

🔗

Which of the following is the best strategy for evaluating

A_{4} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) ?

Compute $A_{4} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20},$ then evaluate the sum.
Rewrite $A_{4} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) = 1 + (- \frac{1}{2} + \frac{1}{2}) + (- \frac{1}{3} + \frac{1}{3}) + (- \frac{1}{4} + \frac{1}{4}) - \frac{1}{5},$ then simplify.

🔗

Activity 8.3.15.

🔗

Recall from Activity 8.3.14 that

{a_{n}}_{n = 1}^{\infty} = {\frac{1}{n} - \frac{1}{n + 1}}

and

A_{n} = \sum_{i = 1}^{n} a_{i} = \sum_{i = 1}^{n} (\frac{1}{i} - \frac{1}{i + 1}) .

🔗

Compute the following partial sums:

$A_{3} .$
$A_{10} .$
$A_{100} .$

🔗

Activity 8.3.16.

🔗

Recall from Activity 8.3.14 that

{a_{n}}_{n = 1}^{\infty} = {\frac{1}{n} - \frac{1}{n + 1}}

and

A_{n} = \sum_{i = 1}^{n} a_{i} = \sum_{i = 1}^{n} (\frac{1}{i} - \frac{1}{i + 1}) .

🔗

Which of the following is equal to

A_{n} ?

$n - \frac{1}{n + 1} .$
$1 - \frac{1}{n} .$
$1 - \frac{1}{n + 1} .$
$1 - \frac{1}{i} .$
$1 - \frac{1}{i + 1} .$

🔗

Definition 8.3.17.

🔗

Given a sequence

{x_{n}}_{1}^{\infty}

and a sequence of the form

{s_{n}}_{1}^{\infty} := {x_{n} - x_{n + 1}}_{1}^{\infty}

we call the series

S_{n} = \sum_{i = 1}^{n} s_{i} = \sum_{i = 1}^{n} (x_{i} - x_{i + 1})

to be a telescoping series.

🔗

Activity 8.3.18.

🔗

Given a telescoping series

S_{n} = \sum_{i = 1}^{n} s_{i} = \sum_{i = 1}^{n} (x_{i} - x_{i + 1}),

find:

$S_{2} .$
$S_{10} .$
Choose $S_{n}$ from the following options:
1. $x_{1} - x_{n}$
2. $x_{1} - x_{n + 1}$
3. $x_{1} - x_{n - 1}$
4. $x_{1} - x_{n} + 1$
5. $x_{1} - x_{n} - 1$

🔗

Activity 8.3.19.

🔗

For each of the following telescoping series, find the closed form for the

n

th partial sum.

$S_{n} = \sum_{i = 1}^{n} (2^{- i} - (2^{- i - 1})) .$
$S_{n} = \sum_{i = 1}^{n} (i^{2} - (i + 1)^{2}) .$
$S_{n} = \sum_{i = 1}^{n} (\frac{1}{2 i + 1} - \frac{1}{2 i + 3}) .$

🔗

Activity 8.3.20.

🔗

Given the closed forms you found in Activity 8.3.19, determine which of the following telescoping series converge. If so, to what value does it converge?

$\sum_{i = 1}^{\infty} (2^{- i} - (2^{- i - 1})) .$
$\sum_{i = 1}^{\infty} (i^{2} - (i + 1)^{2}) .$
$\sum_{i = 1}^{\infty} (\frac{1}{2 i + 1} - \frac{1}{2 i + 3}) .$

🔗

Activity 8.3.21.

🔗

Consider the partial sum sequence

A_{n} = (- 2) + (\frac{2}{3}) + (- \frac{2}{9}) + \dots + (- 2 \cdot {(- \frac{1}{3})}^{n}) .

🔗

(a)

🔗

Find a closed form for

A_{n} .

🔗

(b)

🔗

Does

{A_{n}}

converge? If so, to what value?

🔗

Activity 8.3.22.

🔗

Consider the partial sum sequence

B_{n} = \sum_{i = 1}^{n} (\frac{1}{5 i + 2} - \frac{1}{5 i + 7}) .

🔗

(a)

🔗

Find a closed form for

B_{n} .

🔗

(b)

🔗

Does

{B_{n}}

converge? If so, to what value?

🔗

Subsection 8.3.2 Videos

🔗

Figure 178. Video: Compute the first few terms of a telescoping or geometric partial sum sequence, and find a closed form for this sequence, and compute its limit.

🔗

Subsection 8.3.3 Exercises

🔗

Exercises available at https://tbil.org/calculus/preview/exercises/#/bank/SQ3/

Prev Top Next