Calculus

Infinite Sequences and Series

Infinite Series

Page 1
Problems 1-2
Page 2
Problems 3-7

Example 3.

Show that the harmonic series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{n}\normalsize} \) diverges.

Solution.

To see this, we can write

\[
{\sum\limits_{n = 1}^\infty {\frac{1}{n}} }
= {1 + \frac{1}{2} + \underbrace {\left( {\frac{1}{3} + \frac{1}{4}} \right)}_{\frac{7}{{12}} > \frac{1}{2}} }
+ {\underbrace {\left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right)}_{\frac{{533}}{{840}} > \frac{1}{2}} }+{ \ldots \;\text{and so on}.}
\]

Therefore \(\sum\limits_{n = 1}^\infty {\large\frac{1}{n}\normalsize} \) \(> \sum\limits_{n = 1}^\infty {\large\frac{1}{2}\normalsize} \) = \(\infty .\) Hence, the harmonic series diverges.

Actually, this result was first proved by a mediaeval French mathematician, Nichole Oresme, who lived over \(600\) years ago.

Example 4.

Investigate convergence of the series \(\sum\limits_{n = 0}^\infty {\left( {\large\frac{1}{{{3^n}}}\normalsize} + {\large\frac{1}{{{5^n}}}\normalsize} \right)} .\)

Solution.

This series converges because it is the sum of two convergent series, \(\sum\limits_{n = 0}^\infty {\large\frac{1}{{{3^n}}}\normalsize} \) and \(\sum\limits_{n = 0}^\infty {\large\frac{1}{{{5^n}}}\normalsize}.\) Both are geometric series with ratio \(\left| q \right| \lt 1.\) Then

\[
{\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} }
= {\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}} \right)}^n}} }
= {\frac{1}{{1 – \frac{1}{3}}} }={ \frac{3}{2},}
\]
\[
{\sum\limits_{n = 0}^\infty {\frac{1}{{{5^n}}}} }
= {\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{5}} \right)}^n}} }
= {\frac{1}{{1 – \frac{1}{5}}} }={ \frac{5}{4}.}
\]

Hence, the sum of the given series is

\[
{\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{{3^n}}} + \frac{1}{{{5^n}}}} \right)} }
= {\frac{3}{2} + \frac{5}{4} }={ \frac{{11}}{4}.}
\]

Example 5.

Investigate convergence of the series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}}\normalsize} .\)

Solution.

We see that

\[{\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}} }={ \frac{1}{{n + \pi }} }-{ \frac{1}{{n + \pi + 1}}.}\]

Then the \(n\)th partial sum is

\[
{{S_n} = \left( {\frac{1}{{1 + \pi }} – \frac{1}{{2 + \pi }}} \right) }
+ {\left( {\frac{1}{{2 + \pi }} – \frac{1}{{3 + \pi }}} \right) + \ldots }
+ {\left( {\frac{1}{{n + \pi }} – \frac{1}{{n + \pi + 1}}} \right) }
= {\frac{1}{{1 + \pi }} }-{ \frac{1}{{n + \pi + 1}}.}
\]

Calculate the limit of \({S_n}\) as \(n \to \infty:\)

\[
{\lim\limits_{n \to \infty } {S_n} }
= {\lim\limits_{n \to \infty } \left( {\frac{1}{{1 + \pi }} – \frac{1}{{n + \pi + 1}}} \right) }
= {\frac{1}{{1 + \pi }} }\approx{ 0,24.}
\]

Hence, the series converges.

Example 6.

Determine whether the series

\[{\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} }+{ \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \ldots }+{ \frac{1}{{n\left( {n + 1} \right)}} + \ldots }\]

converges or diverges.

Solution.

The \(n\)th partial sum is

\[{{S_n} = \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} }+{ \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \ldots }+{ \frac{1}{{n\left( {n + 1} \right)}}.}\]

We can easily see that

\[
{\frac{1}{{1 \cdot 2}} = 1 – \frac{1}{2},\;\;}\kern-0.3pt
{\frac{1}{{2 \cdot 3}} = \frac{1}{2} – \frac{1}{3},\;\;}\kern-0.3pt
{\frac{1}{{3 \cdot 4}} = \frac{1}{3} – \frac{1}{4},\;\;}\kern-0.3pt
{\frac{1}{{4 \cdot 5}} = \frac{1}{4} – \frac{1}{5},\;\; \ldots,\;\;}\kern-0.3pt
{{\frac{1}{{n\left( {n + 1} \right)}} }={ \frac{1}{n} – \frac{1}{{n + 1}}.}}
\]

Then

\[
{{S_n} = \left( {1 – \frac{1}{2}} \right) + \left( {\frac{1}{2} – \frac{1}{3}} \right) }
+ {\left( {\frac{1}{3} – \frac{1}{4}} \right) + \ldots }
+ {\left( {\frac{1}{n} – \frac{1}{{n + 1}}} \right).}
\]

Hence,

\[
{{S_n} = 1 – \frac{1}{{n + 1}}\;\;\text{and}\;\;}\kern0pt
{{\lim\limits_{n \to \infty } {S_n} }={ \lim\limits_{n \to \infty } \left( {1 – \frac{1}{{n + 1}}} \right) }={ 1.}}
\]

Thus the series converges to \(1.\)

Example 7.

Evaluate \(\sum\limits_{n = 0}^\infty {\ln {\large\frac{{n + 2}}{{n + 1}}\normalsize}} .\)

Solution.

We write the \(n\)th term as

\[
{{a_n} = \ln \frac{{n + 2}}{{n + 1}} }
= {\ln \left( {n + 2} \right) }-{ \ln \left( {n + 1} \right).}
\]

Calculate the \(n\)th partial sum:

\[
{{S_n} }={ \left( {\ln 2 – \ln 1} \right) }+{ \left( {\ln 3 – \ln 2} \right) }
+ {\left( {\ln 4 – \ln 3} \right) + \ldots }
+ {\left[ {\ln \left( {n + 2} \right) – \ln \left( {n + 1} \right)} \right] }
= {\left( { – \ln 1 + \ln 2} \right) }+{ \left( { – \ln 2 + \ln 3} \right) }
+ {\left( { – \ln 3 + \ln 4} \right) + \ldots }
+ {\left[ { – \ln \left( {n + 1} \right) + \ln \left( {n + 2} \right)} \right] }
= { – \ln 1 + \ln \left( {n + 2} \right) }
= {\ln \left( {n + 2} \right).}
\]

Since \(\lim\limits_{n \to \infty } {S_n} \) \(= \lim\limits_{n \to \infty } \left[ {\ln \left( {n + 2} \right)} \right] \) \(= \infty ,\) we conclude that the given series diverges.

Page 1
Problems 1-2
Page 2
Problems 3-7