Question

Which is the largest number in the following sequence?
${1^{\dfrac{1}{{\sqrt 1 }}}},{3^{\dfrac{1}{{\sqrt 3 }}}},{5^{\dfrac{1}{{\sqrt 5 }}}},...{(2n + 1)^{\dfrac{1}{{\sqrt {2n + 1} }}}},...$
A) ${3^{\dfrac{1}{{\sqrt 3 }}}}$
B) ${5^{\dfrac{1}{{\sqrt 5 }}}}$
C) ${7^{\dfrac{1}{{\sqrt 7 }}}}$
D) The sequence is unbounded

Answer 1

We will start finding the largest number in the sequence by comparing the options given in the question above, here we will see, does the options fulfill the criteria of the sequence. So that is how we will get the answer.

Complete step by step solution:
As given in the question the sequence is
${1^{\dfrac{1}{{\sqrt 1 }}}},{3^{\dfrac{1}{{\sqrt 3 }}}},{5^{\dfrac{1}{{\sqrt 5 }}}},...{(2n + 1)^{\dfrac{1}{{\sqrt {2n + 1} }}}},...$
The criteria for the sequence will be that does it fit in particular number or does it have infinite numbers.
Now we will see the sequence
As
${1^{\dfrac{1}{{\sqrt 1 }}}}$
We will get
$\Rightarrow {1^{\dfrac{1}{1}}} = 1$
As we can see that
$\Rightarrow 3 > 1$
Therefore, on taking square root and taking reciprocal we get,
$\Rightarrow \dfrac{1}{{\sqrt 3 }} > \dfrac{1}{{\sqrt 1 }}$
Similarly, the next number in sequence will be
$\Rightarrow 5 > 3 > 1$
We can clearly see that the sequence is in increasing order
$\Rightarrow \dfrac{1}{{\sqrt 5 }} > \dfrac{1}{{\sqrt 3 }} > \dfrac{1}{{\sqrt 5 }}$
Therefore, If we put any integer in the value of n of the, n does not have particular value and will keep on increasing
${(2n + 1)^{\dfrac{1}{{\sqrt {2n + 1} }}}}$
It will be greater than the previous number. Which will be increasing order sequence is unbounded.

Therefore, the answer to this is D.

Note:
Sequence unbounded means that the sequence is infinite. The sequence doesn’t have an upper limit so it will keep on going without any limit. It is unbounded to the sequence. If a sequence is bounded then it has a lower limit and an upper limit.