Question

Which of the following statements is true?
(a) If a series conditionally converges, then it must absolutely converge as well.
(b) If a sequence \[{\alpha _n}\] converges, then the series \[\sum {{\alpha _n}} \] converges
(c) A sequence which is both bounded and monotonic must converge.
(d) A geometric series converges provided the common ratio is less than \[1\].

Answer 1

Hint: Before answering this question, we need to know some basic definitions that are, conditionally convergent series, absolutely convergent series, convergent sequence, bounded sequence, and monotonic sequence. In order to solve this question, we will analyse the options one by one and check which option satisfies the given condition and hence we get the required result.


Complete step by step answer:
Firstly, let’s recall some definitions
Conditionally convergent series: An infinite series \[\sum\limits_{n = 1}^\infty {{a_n}} \] is said to be converge conditionally if \[\mathop {\lim }\limits_{m \to \infty } \sum\limits_{n = 1}^m {{a_n}} \] exists but \[\sum\limits_{n = 1}^m {|{a_n}|{\text{ }} = \infty } \] .In other words, we can say \[\sum\limits_{n = 1}^\infty {{a_n}} \] is conditionally converges if it converges but not absolutely convergent.
Absolutely convergent series: An infinite series \[\sum\limits_{n = 1}^\infty {{a_n}} \] is said to be converge absolutely if \[\sum\limits_{n = 1}^\infty {|{a_n}|{\text{ }} = L} \] for some real integer \[L\] or we can say an infinite series \[\sum\limits_{n = 1}^\infty {{a_n}} \] is absolutely convergent if the series \[\sum\limits_{n = 1}^\infty {|{a_n}|} \] converges.
Convergent sequence: A sequence is said to be convergent if any sequence \[{S_n}\] converges to a limit \[S\] i.e., \[\mathop {\lim }\limits_{n \to \infty } {S_n} = S\] if, for any \[\varepsilon > 0\] , there exists an \[N\] such that \[|{S_n} - S|{\text{ }} < \varepsilon \] for \[n > N\]
Bounded sequence: A sequence is said to be bounded if it is both bounded above and bounded below.
A sequence is bounded above if \[\exists \] a number \[N\] such that \[{a_n} \leqslant N\] for every \[n \geqslant 1\]
A sequence is bounded below if \[\exists \] a number \[M\] such that \[{a_n} \geqslant M\] for every \[n \geqslant 1\]
Monotonic sequence: A sequence is said to be a monotonic sequence if it is increasing or decreasing.
A sequence \[\left\{ {{a_n}} \right\}\] is called increasing if \[{a_n} \leqslant {a_{n + 1}}{\text{ }}\forall n \in \mathbb{N}\]
A sequence \[\left\{ {{a_n}} \right\}\] is called decreasing if \[{a_n} \geqslant {a_{n + 1}}{\text{ }}\forall n \in \mathbb{N}\]
Now, let’s analyse the options one by one.
Option (a) says: If a series conditionally converges, then it must absolutely converge as well.
As from the definition of conditionally convergent series, a series is said to be conditionally convergent if it is convergent but not absolutely convergent.
For example: the alternating harmonic series i.e., \[\sum\limits_{n = 1}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}}}{n}} \] converges but if we take absolute value of each term, then a standard harmonic series is formed i.e., \[\sum\limits_{n = 1}^\infty {|\dfrac{{{{\left( { - 1} \right)}^n}}}{n}} | = \sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] which is an example of a divergent series. Hence, the given statement is false.
Now, option (b) says: If a sequence \[{\alpha _n}\] converges, then the series \[\sum {{\alpha _n}} \] converges
The statement is not true. For example- let’s consider the sequence, \[\left\{ {{a_n}} \right\} = \left\{ {\dfrac{1}{n}} \right\}\] which is convergent means \[\mathop {\lim }\limits_{n \to \infty } \left\{ {{a_n}} \right\} = \mathop {\lim }\limits_{n \to \infty } \left\{ {\dfrac{1}{n}} \right\} = 0\] but if we take summation i.e., \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] which is a harmonic series and is divergent. Hence, the given statement is false.
Now, option (c) says: A sequence which is both bounded and monotonic must converge.
This statement is true. Because there is a well define theorem which states that if \[\left\{ {{a_n}} \right\}\] is bounded and monotonic, then \[\left\{ {{a_n}} \right\}\] is convergent. Hence, the given statement is true.
Now, there is no need to check option (d) as we get our required result by the above statement only.

Note: We can show or verify out answer with the help of an example, let us consider the sequence \[\left\{ {{a_n}} \right\}\] defined as follows:
\[{a_1} = 2\]
\[{a_{n + 1}} = \dfrac{{{a_n} + 5}}{3}\] for \[n \geqslant 1\]
First, we will show that the sequence is monotonically increasing.
Since \[{a_2} = {a_{1 + 1}} = \dfrac{{{a_1} + 5}}{3} = \dfrac{7}{3} > 2 = {a_1}\]
\[{a_3} = {a_{2 + 1}} = \dfrac{{{a_2} + 5}}{3} = \dfrac{{\dfrac{7}{3} + 5}}{3} = \dfrac{{22}}{9} > \dfrac{7}{3} = {a_2}\]
Proceeding in the same way, we get
\[{a_{n + 1}} \geqslant {a_n}\]
Hence, the sequence is monotonically increasing.
Now, we prove that the sequence is bounded by \[3\] .The statement is clearly true for \[n = 1\] .Suppose \[{a_k} \leqslant 3\] for some \[k \in \mathbb{N}\] .Then
\[{a_{k + 1}} = \dfrac{{{a_k} + 5}}{3} \leqslant \dfrac{{3 + 5}}{3} = \dfrac{8}{3} \leqslant 3{\text{ }} - - - \left( 1 \right)\]
It follow that \[{a_n} \leqslant 3{\text{ }}\forall n \in \mathbb{N}\]
Now, from the monotone convergence theorem,
We deduce that there is \[l \in \mathbb{R}\] such that \[\mathop {\lim }\limits_{n \to \infty } {a_n} = l\] .Since the subsequence \[\left\{ {{a_{k + 1}}} \right\}_{k = 1}^\infty \] also converges to \[l\] ,so taking limits in the equation \[\left( 1 \right)\] we get
\[l = \dfrac{{l + 5}}{3}\]
Therefore, \[3l = l + 5\]
Hence, \[l = \dfrac{5}{2}\] which means the sequence converges.
Also remember, the converse of the option (a) and (b) is true while the converse of the option (C) may or may not be true.