# Understand the problem

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$a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$   Which of the following is true? (a) sup {$$a_n|n \in \mathbb{N}$$}=3 and inf {$$a_n|n \in \mathbb{N}$$}=1 (b) lim inf ($$a_n$$)=lim sup ($$a_n$$)=$$\frac{3}{2}$$ (c) sup {$$a_n|n \in \mathbb{N}$$}=2 and inf {$$a_n|n \in \mathbb{N}$$}=1 (d) lim inf ($$a_n$$)= 1 lim sup ($$a_n$$)=3

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]IIT JAM 2018 Question 11[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Infinum and Supremum[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.1" open="off"]Real Analysis By S K Mapa[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" custom_padding="|||44px||"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"]

$a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Now the limit points of this set are those points which the set does not attain.So, they might be the sup and inf which are not attained by this set. Basically sup($$a_n$$)= max{ limit points, $$a_n$$ | n $$\in$$ $$\mathbb{N}$$} Limit points are $$2,1$$ and $$a_1= 2+1=3, a_3= 2- \frac{1}{3} ; a_5= 2+\frac{1}{5}$$ $$a_0= 1+1=2 , a_2= 1+ \frac{1}{4} , a_3= 1+\frac{1}{8}$$ Now you can calculate  the supremum?

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]From the observation of Hint 2 we have  sup  $$a_n$$= max $$\{2,1,3,2\}=3$$ Similarly, inf $$a_n$$= min$$\{$$ limit points, $$a_n | n \in \mathbb{N}\}$$ Can you calculate that by yourself?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]

inf $$a_n$$= min {2,1,2 -$$\frac{1}{3}$$}=1 So, option A is correct. Now there is another question regarding  lim sup and lim inf. We can observe that we have mainly $$3$$  subsequences , corresponding to  $$n$$ is even; $$n=2k$$ $$n$$= $$4k+1$$ $$n=4k+3$$

Can you calculate the corresponding subsequences  and their limits?

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1"]For $$n=2k$$ we have $$a_{2k}=1+ \frac{1}{2^{ek}} \longrightarrow 1$$ ask For $$a_{4k+1}= 2+ \frac{1}{4k+1} \longrightarrow 2$$ ask $$a_{4k+3}= 2-\frac{1}{4k+3} \longrightarrow 2$$ ask So, lim sup $$a_n$$=max$$\{1,2\}=2$$ Lim inf $$a_n$$=min$$\{1,2\}=1$$ Therefore, Option C is also correct[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

# Look At The Knowledge Graph

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# Similar Problems

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# What are we learning?

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" text_font_size="18px" background_color="#f4f4f4" custom_margin="50px||50px||false|false" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Sequences & Subsequences are the key features in the filed of real analysis. We will see how to imply these concepts in our problem

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="50px||50px||false|false" custom_padding="10px|20px|10px|20px" hover_enabled="0" box_shadow_style="preset2"]Let $$s_n$$ = 1+$$\frac{1}{1!}$$+$$\frac{1}{2!}$$+........+$$\frac{1}{n!}$$ for n $$\in$$ $$\mathbb{N}$$ Then which of the following is TRUE for the sequence {$$s_{n}\}^\infty_{n=1}$$:   (a) {$$s_{n}\}^\infty_{n=1}$$ converges in $$\mathbb{Q}$$.   (b) {$$s_{n}\}^\infty_{n=1}$$ is a Cauchy sequence but does not converges to $$\mathbb{Q}$$.   (c) The subsequence  {$$s_{k^n}\}^\infty_{n=1}$$ is convergent in $$\mathbb{R}$$ when k is a even natural number.   (d) {$$s_{n}\}^\infty_{n=1}$$ is not a Cauchy sequence. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0.9" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0.9"]IIT Jam 2018[/et_pb_accordion_item][et_pb_accordion_item title="Key competency" _builder_version="4.0.9" open="off"]Gradient[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]
 Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2 Tom M. Apostol

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" height="436px" max_height="720px"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0.9"]

I am going to give you 3 clues in the beginning you try to work out using them. Then I will elaborate this clues in the following hints  (I) Every convergent sequence is a Cauchy sequence  (II)Every subsequence of a convergent sequence is convergent  (III)Consider then term 1+$$\frac{1}{1!}$$+$$\frac{1}{2!}$$+........+$$\frac{1}{n!}$$ Does this remind you any well known series?

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I wil start with (III) consider $$e^x$$=1+$$\frac{x}{1!}$$+$$\frac{x^2}{2!}$$+........+$$\frac{x^n}{n!}$$ Isn't the seris that we have to , is the value at x=1. Hence the given series$$\rightarrow$$ e $$\in$$ $$\mathbb{R}$$ \ $$\mathbb{Q}$$

So option (a) is incorrect.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]

Every subsequence of a convergent sequence is convergent so {$$s_{k^n}\}^\infty_{n=1}$$ is convergent not only for even k, but for any $$k \in \Bbb N$$. So option (c) is incorrect.

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Every convergent sequence is a Cauchy sequence so option (d) is incorrect and $$e \in$$ $$\mathbb{R}$$ so the given subsequence is convergent in $$\mathbb{R}$$. So only option (b) is correct.

# Fun fact: Do you know that this man has a sequence named after him?

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