# 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.# Take A Look Into This Knowledge Graph

# Connected Program at Cheenta

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

# Similar Problems

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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 $$\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b}\left(\log _{e} n\right)^{c}} ?$$
(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.# Take A Look Into This Knowledge Graph

# Connected Program at Cheenta

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

# Similar Problems

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Thank you for help

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