Exercises: Series

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\bigmath }[1]{$\displaystyle #1$} \newcommand {\choicebreak }[0]{} \newcommand {\type }[0]{} \newcommand {\notes }[0]{} \newcommand {\keywords }[0]{} \newcommand {\offline }[0]{} \newcommand {\comments }[0]{\begin {feedback}} \newcommand {\multiplechoice }[0]{\begin {multipleChoice}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Exercises relating to fundamental properties of series.

Compute the sum of the series:

Split the series into two separate geometric series and evaluate each separately.

$\sum _{n=0}^\infty \frac {3^n - 2^n}{4^n} = \answer {2}.$

Compute the sum of the series:

One way to proceed is to reindex the series so that it starts at $n = 0$ .

$\sum _{n=3}^\infty \frac {(-1)^{n+1} 2^n}{3^n} = \answer { \frac {8}{45} }$

Compute the sum of the series: $\sum _{n=1}^\infty \frac {3^{2n+1}}{2^{3n} \cdot 9^n} = \answer {\frac {3}{7}}$

Simplify as much as possible and write in standard form for a geometric series.

Give an example of a geometric series whose first term is $3$ and which sums to $6$ . If no such series exists, enter N/A in both spaces below: $6 = \sum _{n=0}^\infty \answer {3} \left ( \answer {\frac {1}{2}} \right )^n$
Give an example of a geometric series whose first term is $3$ and which sums to $2$ . If no such series exists, enter N/A in both spaces below: $2 = \sum _{n=0}^\infty \answer {3} \left ( \answer {- \frac {1}{2}} \right )^n$
Give an example of a geometric series whose first term is $3$ and which sums to $1$ . If no such series exists, enter N/A in both spaces below: $1 = \sum _{n=0}^\infty \answer {N/A} \left ( \answer {N/A} \right )^n$

Find the correct value of the constant in the blank below which makes the sum of the series equal to zero: $0 = \sum _{n=1}^\infty \frac {(-1)^{n-1}\answer {8} - 2^n}{3^n}$

One reasonable strategy is to separate the series into two geometric series and reindex to apply the standard formula.

Compute the sum of the infinite series below: $\sum _{n=1}^\infty \frac {1}{n(n+1)(n+2)} = \answer {\frac {1}{4}}.$

There is a closely-related telescoping series: $\frac {1}{n(n+1)} - \frac {1}{(n+1)(n+2)} = \frac {2}{n(n+1)(n+2)}$

Compute the sum of the infinite series below. $\sum _{n=1}^\infty \frac {2}{n(n+2)} = \answer {\frac {3}{2}}$

Expand the terms using partial fractions and compute several partial sums by hand. What you get is something like a telescoping series, but cancellations occur in a slightly different way than usual.

$\frac {2}{n(n+2)} = \frac {1}{n} - \frac {1}{n+2}.$

The formula for a general partial sum is $\frac {2}{1 \cdot 3} + \cdots + \frac {2}{N \cdot (N+2)} = 1 + \frac {1}{2} - \frac {1}{N+1} - \frac {1}{N+2}$

Use your answer to the previous exercise to compute the sum of the series: $\sum _{n=3}^\infty \frac {1}{n(n-2)} = \answer {\frac {3}{4}}$
Use the answer you just found to compute the sum of the series: $\sum _{n=4}^\infty \frac {1}{n(n-2)} = \answer {\frac {5}{12}}$
To drop the $n=3$ term, you can just calculate it and then subtract it from the previous series.

Sample Quiz Questions

Compute the exact value of the infinite series $\sum _{n=1}^\infty \ln \left ( \frac {1 + n^{-1}}{1 + (n+1)^{-1}} \right ).$

$\ln 2$ $\ln 3$ $\ln 4$ $\ln 5$ $\ln 6$ $\ln 7$

The series is not a geometric series or Taylor series, we compute the first few partial sums: $\begin {aligned} S_1 & = \ln \left ( \frac {1 + 1}{1 + 2^{-1}} \right ) = \ln \left ( \frac {2}{\frac {3}{2}} \right ) \\ S_2 & = \ln \left ( \frac {1 + 1}{1 + 2^{-1}} \right ) + \ln \left ( \frac {1 + 2^{-1}}{1 + 3^{-1}} \right ) = \ln \left ( \frac {1 + 1}{1 + 3^{-1}} \right ) = \ln \left ( \frac {2}{\frac {4}{3}} \right ) \\ S_3 & = \ln \left ( \frac {2}{1 + 3^{-1}} \right ) + \ln \left ( \frac {1 + 3^{-1}}{1 + 4^{-1}} \right ) = \ln \left ( \frac {2}{1 + 4^{-1}} \right ) = \ln \left ( \frac {2}{\frac {5}{4}} \right ) \\ & \ \vdots \\ S_n & = \ln \left ( \frac {2}{1 + (n+1)^{-1}} \right ). \end {aligned}$ In particular, writing the sum of logarithms as a logarithm of a product leads to substantial cancellation. By letting $n \rightarrow \infty$ , we get $S_n \rightarrow \ln 2$ .