Test of Mathematics Solution Subjective 155 -The Lim 1/(n+r) Problem

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 155 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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Problem

  Evaluate: $ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n})$

Solution

As the title suggests the modification of this problem will be, that we will solve a more general series and then use a specific value to arrive at the solution of this problem.

First let us consider the following limit:

$ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+kn})$

Observe carefully that using k=1 in this limit, we get the limit that has been asked to evaluate.

Now

$ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}) = \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{1}{n+r})$

$ = \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{\frac{1}{n}}{1+\frac{r}{n}})$

$ = \lim_{n\to\infty}\frac{1}{n} (\sum_{r=1}^{kn} \frac{1}{1+\frac{r}{n}})$

Let's substitute $ \frac{r}{n} = x  =>  dr = ndx$

Now we can change the sum to an integral

$ => \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{\frac{1}{n}}{1+\frac{r}{n}}) = \lim_{n\to\infty} \frac{1}{n}*n\int_{0}^{k} \frac{1}{1+x} dx $

$ = \lim_{n\to\infty}( log |x+1|_{k} -  log |x+1|_{0})$

$ = \lim_{n\to\infty} log |k+1|$

$ = log |k+1|$                     (As the term is an 'n' free term)

So we see the solution is $ = log |k+1|$ 

Substituting k=1, we get 

$ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}) = \log {2}$ 

CLT and Confidence Limits | ISI MStat 2016 PSB Problem 8

This is a problem from ISI MStat Examination 2016. This primarily tests the student's knowledge in finding confidence intervals and using the Central Limit Theorem as a useful approximation tool.

The Problem:

Let \( (X_1,Y_1),(X_2,Y_2),...,(X_n,Y_n) \) be independent and identically distributed pairs of random variables with \( E(X_1)=E(Y_1) \) , \(\text{Var}(X_1)=\text{Var}(Y_1)=1 \), and \( \text{Cov}(X_1,Y_1)= \rho \in (-1,1) \)

(a) Show that there exists a function \( c(\rho) \) such that :

\(\lim_{n \rightarrow \infty} P(\sqrt{n}(\bar{X}-\bar{Y}) \le c(\rho) )=\Phi(1)\) , where \( \Phi \) is the cdf of \( N(0,1) \)

(b)Given \( \alpha>0 \) , obtain a statistic \(L_n \) which is a function of \( (X_1,Y_1),(X_2,Y_2),..,(X_n,Y_n) \) such that
\(\lim_{n \rightarrow \infty} P(L_n<\rho<1)=\alpha \) .

Prerequisites:

(a)The Central Limit Theorem and Convergence of Sequence of RVs

(b)Idea of pivots and how to obtain confidence intervals.

Solution:

(a) See that \( E(\bar{X}-\bar{Y})=0 \).

See that \( \text{Var}(\bar{X}-\bar{Y})=\frac{\sum_{i=1}^{n} \text{Var}(X_i-Y_i)}{n^2} =\frac{2(1-\rho)}{n} \)

Take \( c(\rho)=2(1-\rho) \).

By the Central Limit Theorem,

see that \( P(\frac{\sqrt{n}(\bar{X}-\bar{Y})}{\sqrt{2(1-\rho)}} \le 1) \rightarrow \Phi(1) \) as \(n \rightarrow \infty \)

(b) Let \( Z_i = X_i - Y_i \)

\(Z_i\)'s are iid with \( E(Z_i)=0 \) and \(V(Z_i)=2(1- \rho) \)

Again, by CLT, \( \frac{\sqrt{n} \bar{Z}}{\sqrt{2-2\rho}} \stackrel{L}\longrightarrow N(0,1) \)

Use this as the pivot to obtain an asymptotic confidence interval for \( \rho \).

See that \( P(\frac{\sqrt{n} \bar{Z}}{\sqrt{2-2\rho}}) \ge \tau_{\alpha})=\alpha \) , where \( \tau_\alpha \) : upper \(\alpha\) point of \(N(0,1) \).

Equivalently , you can write, \( P( \rho \ge 1- \frac{n \bar{Z}^2 }{2 \tau_{\alpha}^2} ) =\alpha \) , as \( n \rightarrow \infty \).

Thus , \(L_n=1- \frac{n \bar{Z}^2 }{2 \tau_{\alpha}^2} \).

Food For Thought:

Suppose \(X_1,..,X_n \) are equi-correlated with correlation coefficient \( \rho \).

Given, \( E(X_i)= \mu \) , \(V(X_i) = \sigma_i ^2 \), for \(i=1,2,..,n\).

Can you see that \( \rho \ge -\frac{1}{\frac{(\sum \sigma_i)^2}{\sum \sigma_i ^2} -1 } \) ?

It's pretty easy right? Yeah, I know 😛 . I am definitely looking forward to post inequalities more often .

This may look trivial but it is a very important result which can be used in problems where you need a non-trivial lower bound for \( \rho \) .

Well,well suppose now \(Y_i=X_i - \bar{X} \).

Can you show that a necessary and sufficient condition for \(X_i\) , \( \{ i=1,2,..,n \} \) to have equal variance is that \( Y_i \) should be uncorrelated with \( \bar{X} \)?

Till then Bye!

Stay Safe.

Radius of Convergence of a Power series | IIT JAM 2016

Try this problem from IIT JAM 2017 exam (Problem 48) and know how to determine radius of convergence of a power series.

Radius of Convergence of a Power Series | IIT JAM 2016 | Problem 48

Find the radius of convergence of the power series
$$
\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}
$$

Key Concepts

Real Analysis

Series of Functions

Power Series

Check the Answer

Answer: $\frac12$

IIT JAM 2016 , Problem 48

Real Analysis : Robert G. Bartle

Try with Hints

Given, the power series is $\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}$.

Let us put $2n=m$ to get the standard form of a power series.

We get,

$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(x+2)^{ m}$.

Now let us make the transformation $z=x+2$ to get a power series about 0 :

We have,

$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(z)^{ m}$

Compairing with $ \sum_{m=2}^{\infty} a_m (z)^m$

we get,

$a_m= \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)} $

Now we have to test the convergence of the series.

Can you apply Ratio Test to check the convergence of the series.

Ratio Test : Let $\sum_{n=0}^{\infty} a_{n} x^{n}$ be a power series and let $\lim \left|\frac{a_{n+1}}{a_{n}}\right|=\mu .$ Then

  1. if $\mu=0$ the series is everywhere convergent;
  2. if $0<\mu<\infty$ the series is absolutely convergent for all $x$ satisfyir $|x|<\frac{1}{\mu}$ and the series is divergent for all $x$ satisfying $|x|>\frac{1}{\mu}$
  3. if $\mu=\infty,$ the series is nowhere converegnt.

$\begin{aligned}\left|\frac{a_{m+1}}{a_m}\right| &=\left| \frac{4^{\frac m2}\cdot 2\cdot4}{(m+1)(m+3)} \times \frac{m(m+2)}{4^{\frac m2} \cdot 4}\right| \\&=\left| \quad \frac{2\left(1+\frac{2}{m}\right)}{\left(1+\frac{1}{m}\right)(1+\frac 3m)}\right|\end{aligned}$

Now

$\lim \left|\frac{a_{m+1}}{a_{m}}\right|=2 \in (0,\infty)$

Then, The given power series is absolutely convergent i.e., convergent $\forall x$ such that $|x+2|<\frac 12$

Then the answer is $\frac 12$

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Series Convergence: IIT JAM 2018 Problem 12

Understand the problem

Let $a, b, c \in \mathbb{R} .$ Which of the following values of $a, b, c$ do NOT result in the convergence of the series $$\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b}\left(\log _{e} n\right)^{c}} ?$$ (A) $|a|<1, b \in \mathbb{R}, c \in \mathbb{R}$ (B) $a=1, b>1, c \in \mathbb{R}$ (C) $a=1, b \geq 0, c<1$ (D) $a=-1, b \geq 0, c>0$  

Start with hints

[Hint 1] One disclaimer: In this question you will  see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to  give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz's test states that if ${u_n}$ be a monotone decreasing sequence of positive real numbers and lim $u_n = 0$ , then the alternating series $u_1 - u_2 + u_3 - u_4 + ...........$ is convergent.] [Hint 2] $\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} $ Let us talk about option D first $a=-1, b \geq 0 , c<0 $ $\sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}} $  where $\frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 $ as $n \longrightarrow \infty $ Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence $f(n) $ of non-negative real numbers, the series $\sum_{n=1}^{\infty} f(n) $ converges if and only if the "condensed series" $\sum_{n=0}^{\infty} 2^{n} f(2^{n})$ converges. Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let $\sum u_n $ be a series of positive real numbers and let $lim \frac{u_n +1}{u_n} = l $ Then $\sum u_n $ is convergent if $l<1 $ , $\sum u_n $ is divergent if $ l>1 $. [Hint 3]Moving on to option c $ a=1 , b \geq 0 , c<1 $ $ \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S $(say) Observe if we have $c=b=\frac{1}{3} $ thus $S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty $ So, by comparison test(Let $\sum u_n $ and $\sum v_n $ be two series of positive real numbers and there is a natural number m such that $u_n \leq kv_n $ for all $n \geq m,k $ being a fixed positive number. Then (i)   $\sum u_n $ is convergent if $\sum v_n $ is convergent.  we have S is divergent. (ii) $ \sum u_n $ is divergent if $\sum u_n $ is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making $ b>1 $ say $b=2 $ and make $x c $ smaller  Let $ c= \frac{1}{2} $ Thus $S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} $ Here $\sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty $ So, $ S $ is convergent and c is one correct answer. Look for the others. [Hint 4]Option b $ a=1,b>1,c \in \mathbb{R}$ $S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}$ If $ c=2 $ clearly by comparison test $S $ will be convergent  Now the question is that, can we find one example such that the series will be divergent? Observe that, if $ c \geq 0 $ then as $b>1 $ we will get that the series is convergent. What will happen if $c<0 $  Here Cauchy Condensation test comes into play  Consider $a=2>1 $ thus $\sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}} =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}} =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}}$ Now we have to use D’ Alembert’s Ratio test : Consider $ a_n = \frac{1}{(2^{b-1})^{n} n^{c}} $ Thus $\frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1 $ Hence the series is convergent and so the series is convergent for any value of $c $ and here the series is convergent always and that is why option b is not correct. [Hint 5]option a) $ |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} $ Here if we consider $ a<0 , b<0 , c<0 $ The series is convergent by Leibnitz test . So, the question is whether we can find out some values of $a,b,c $ such that the series will be divergent. Consider $ a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} $ $ \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}$  Now we know that  $ \frac{n+1}{n} \longrightarrow 1 $ We have to think about $ \frac{\log(n+1)}{\log n}$ Let us consider $ \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1$[using L'Hopital's rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if  $ {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 $ or $ -\infty , +\infty , g'(x) \neq 0 $ for all x in I with $ x \neq c$ and ${\lim}_{x \to c} \frac{f'(x)}{g'(x)}$ exist then ${\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}$  ] So, $\frac{\log(n+1)}{\log n} \to 1$ And hence $ \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1$. So,the series is convergent $\forall |a|<1,b,c \in \mathbb{R} $Hence c. is the only correct answer.

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Warm Yourself With An Mcq

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let a, b, c $latex \in \mathbb{R} $  Which of the following values of a ,b, c do NOT result in the convergence of the series  $latex \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_en)^{c}} $ (a) $latex |a|<1 , b \in \mathbb{R} , c \in \mathbb{R} $ (b) $latex a=1 , b>1 , c \in \mathbb{R} $ (c) $latex a=1 , b \leq 1 , c<1 $ (d) $latex a=-1 , b \geq , c>0 $  

[/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="4.1" 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 Problem 12 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Convergence of a seris [/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="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#8300e9" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/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"][et_pb_tab title="Hint 0" _builder_version="4.1"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"] One disclaimer: In this question you will  see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to  give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz's test states that if $latex {u_n}$ be a monotone decreasing sequence of positive real numbers and lim $latex u_n = 0$ , then the alternating series $latex u_1 - u_2 + u_3 - u_4 + ...........$ is convergent.][/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"] $latex \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} $ Let us talk about option D first $latex a=-1, b \geq 0 , c<0 $ $latex \sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}} $  where $latex \frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 $ as $latex n \longrightarrow \infty $ Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence $latex f(n) $ of non-negative real numbers, the series $latex \sum_{n=1}^{\infty} f(n) $ converges if and only if the "condensed series" $latex \sum_{n=0}^{\infty} 2^{n} f(2^{n})$ converges.Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let $latex \sum u_n $ be a series of positive real numbers and let $latex lim \frac{u_n +1}{u_n} = l $ Then $latex \sum u_n $ is convegent if $latex l<1 $ , $latex \sum u_n $ is divergent if $latex l>1 $.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]Moving on to option c $latex a=1 , b \geq 0 , c<1 $ $latex \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S $(say) Observe if we have $latex c=b=\frac{1}{3} $ thus $latex S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty $ So, by comparison test(Let $latex \sum u_n $ and $latex \sum v_n $ be two series of positive real numbers and there is a natural number m such that $latex u_n \leq kv_n $ for all $latex n \geq m,k $ being a fixed positive number. Then (i)   $latex \sum u_n $ is convergent if $latex \sum v_n $ is convergent.  we have S is divergent. (ii) $latex \sum u_n $ is divergent if $latex \sum u_n $ is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making $latex b>1 $ say $latex b=2 $ and make $latex c $ smaller  Let $latex c= \frac{1}{2} $ Thus $latex S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} $ Here $latex \sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty $ So, $latex S $ is convergent and c is one correct answer. Look for the others.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1"]Option b $latex a=1,b>1,c \in \mathbb{R}$ $latex S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}$ If $latex c=2 $ clearly by comparison test $latex S $ will be convergent  Now the question is that, can we find one example such that the series will be divergent? Observe that, if $latex c \geq 0 $ then as $latex b>1 $ we will get that the series is convergent. What will happen if $latex c<0 $  Here Cauchy Condensation test comes into play  Consider $latex a=2>1 $ thus $latex \sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}} =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}} =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}}$ Now we have to use D’ Alembert’s Ratio test : Consider $latex a_n = \frac{1}{(2^{b-1})^{n} n^{c}} $ Thus $latex \frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1 $ Hence the series is convergent and so the series is convergent for any value of $latex c $ and here the series is convergent always and that is why option b is not correct.[/et_pb_tab][et_pb_tab title="Hint 5" _builder_version="4.1"]option a) $latex |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} $ Here if we consider $latex a<0 , b<0 , c<0 $ The series is convergent by Leibnitz test . So, the question is whether we can find out some values of $latex a,b,c $ such that the series will be divergent. Consider $latex a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} $ $latex \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}$  Now we know that  $latex \frac{n+1}{n} \longrightarrow 1 $ We have to think about $latex \frac{\log(n+1)}{\log n}$ Let us consider $latex \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1$[using L'Hopital's rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if  $latex {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 $ or $latex -\infty , +\infty , g'(x) \neq 0 $ for all x in I with $latex x \neq c$ and $latex {\lim}_{x \to c} \frac{f'(x)}{g'(x)}$ exist then $latex {\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}$  ] So, $latex \frac{\log(n+1)}{\log n} \to 1$ And hence $latex \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1$. So,the series is convergent $latex \forall |a|<1,b,c \in \mathbb{R} $Hence c. is the only correct answer.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#e06100" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" hover_enabled="0" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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TIFR 2013 Problem 17 Solution -Convergence of Improper Integral


TIFR 2013 Problem 17 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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Problem:True/False

True/ False?

The integral \(\int_{0}^{\infty} e^{-x^5}dx \) is convergent.

Hint:

As x varies from 0 to "infinity", \(-x^5\) varies from 1 to "minus infinity". Basically, the function is "rapidly decreasing". We can hope that nothing goes wrong in the convergence.

Discussion:

Recall that \(\int_{0}^{\infty} e^{-x}dx \) is convergent. (In fact the value of this integral is 1 which can be done by simple calculation). We try to use this for our comparison test. For \(1\le x < \infty \)   \(e^{-x^5} \le e^{-x} \). This is because in this interval, \(x^5 \ge x\) and because \(e^x\) is an increasing function. So for \(1\le x < \infty \), our worries are over.

\(\int_{1}^{\infty} e^{-x^5} \le \int_{1}^{\infty} e^{-x}dx < \infty \).

Now we only need to check whether the integration is finite on \([0,1]\). The intergand is continuous and bounded by 1 (because it is monotonic decreasing and value at 0 is 1) on this interval, hence the integral is bounded. \(\int_{0}^{1} e^{-x^5}dx \le \int_{0}^{1} 1dx = 1 \).

Thus, \(\int_{0}^{\infty} e^{-x^5}dx \) is convergent.

Helpdesk

TIFR 2013 problem 15 | Problem on Convergence

Try this problem based on convergence from TIFR 2013, Problem 15.

Question: TIFR 2013 problem 15

True/False?

Let \(x_1\in(0,1)\). For \(n>1\) define \(x_{n+1}=x_n-x_n^{n+1} \) Then \(lim_{n\to \infty} x_n \) exists.

Hint: A bounded monotone sequence is convergent.

Discussion:

By induction, we can prove that the sequence is between 0 and 1 always.

Suppose, \(x_n\in(0,1)\). Since we are removing a positive number from \(x_n\) to get \(x_{n+1}\), we have \(x_{n+1}<x_n<1\). This also shows that the sequence is decreasing. And since for a number in between 0 and 1, when we take positive powers it decreases, in other words \(x_n>x_n^{n+1}\) we have \(x_{n+1}>0\).

Therefore the given sequence is decreasing and bounded below by 0. Hence it is convergent.

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