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Series Convergence: IIT JAM 2018 Problem 12

Well this problem appeared in IIT JAM 2018 paper . This problem is very basic one and requires basic knowledge form convergence of a seris.

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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} (log_en)^{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 \leq 1 , c<1(d) a=-1 , b \geq , c>0 
Source of the problem
IIT JAM 2018 Problem 12 
Topic
Convergence of a seris 
Difficulty Level
Easy
Suggested Book
Real Analysis By S.K Mapa

Start with hints

Do you really need a hint? Try it first!
 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.]
 \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 0as n \longrightarrow \inftyHence 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_nbe a series of positive real numbers and let lim \frac{u_n +1}{u_n} = lThen \sum u_nis convegent if l<1, \sum u_nis divergent if l>1.
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 \inftySo, by comparison test(Let \sum u_nand \sum v_nbe two series of positive real numbers and there is a natural number m such that u_n \leq kv_nfor all n \geq m,kbeing a fixed positive number. Then (i)   \sum u_nis convergent if \sum v_nis convergent.  we have S is divergent. (ii) \sum u_nis divergent if \sum u_nis divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making b>1say b=2and make csmaller  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}} < \inftySo, Sis convergent and c is one correct answer. Look for the others.
Option b a=1,b>1,c \in \mathbb{R}S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}If c=2clearly by comparison test Swill be convergent  Now the question is that, can we find one example such that the series will be divergent? Observe that, if c \geq 0then as b>1we will get that the series is convergent. What will happen if c<0  Here Cauchy Condensation test comes into play  Consider a=2>1thus \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}} < 1Hence the series is convergent and so the series is convergent for any value of cand here the series is convergent always and that is why option b is not correct.
option a) |a| < 1, b \in \mathbb{R} , c \in \mathbb{R}Here if we consider a<0 , b<0 , c<0The series is convergent by Leibnitz test . So, the question is whether we can find out some values of a,b,csuch 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 1We 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) = 0or -\infty , +\infty , g'(x) \neq 0for all x in I with x \neq cand {\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 1And 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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2 replies on “Series Convergence: IIT JAM 2018 Problem 12”

Suggest some good questions for exam oriented questions for the IIT JAM for sum of series and radius of convergence

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