# Infinite Series – Page 2

• ### 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