Infinite Series
Solved Problems
Example 3.
Show that the harmonic series \[\sum\limits_{n = 1}^\infty {\frac{1}{n}} \] diverges.
Solution.
To see this, we can write
\[\sum\limits_{n = 1}^\infty {\frac{1}{n}} = 1 + \frac{1}{2} + \underbrace {\left( {\frac{1}{3} + \frac{1}{4}} \right)}_{\frac{7}{{12}} \gt \frac{1}{2}} + \underbrace {\left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right)}_{\frac{{533}}{{840}} \gt \frac{1}{2}} + \ldots \;\text{and so on}.\]
Therefore \(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) \(\gt \sum\limits_{n = 1}^\infty {\frac{1}{2}} \) = \(\infty .\) 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 \[\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{{3^n}}}} + {\frac{1}{{{5^n}}}} \right)}.\]
Solution.
This series converges because it is the sum of two convergent series, \(\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \) and \(\sum\limits_{n = 0}^\infty {\frac{1}{{{5^n}}}}.\) 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 \[\sum\limits_{n = 1}^\infty {\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}}}.\]
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
\[\frac{1}{{1 \cdot 2}} = 1 - \frac{1}{2},\;\; \frac{1}{{2 \cdot 3}} = \frac{1}{2} - \frac{1}{3},\;\; \frac{1}{{3 \cdot 4}} = \frac{1}{3} - \frac{1}{4},\;\; \frac{1}{{4 \cdot 5}} = \frac{1}{4} - \frac{1}{5},\;\; \ldots,\;\; \frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}.\]
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,
\[{S_n} = 1 - \frac{1}{{n + 1}}\;\;\text{and}\;\; \lim\limits_{n \to \infty } {S_n} = \lim\limits_{n \to \infty } \left( {1 - \frac{1}{{n + 1}}} \right) = 1.\]
Thus the series converges to \(1.\)
Example 7.
Evaluate \[\sum\limits_{n = 0}^\infty {\ln {\frac{{n + 2}}{{n + 1}}}}.\]
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.
