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Question: Find the range of $x$ for the convergence of the series$$\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$$ kjn

MY Approach: By $n$th term divergence test, $$\lim_{n\rightarrow\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}=0 \Longleftrightarrow |4x-12|<1\Longleftrightarrow\frac{11}{4}<x<\frac{13}{4},$$ not in the options.

Edit

Cauchy Root Test

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    $\begingroup$ If $\lim_{n\to\infty} a_n = 0$, that does not mean that the sum $$\sum_{n=1}^\infty a_n$$ converges. $\endgroup$ – 5xum Jan 24 '18 at 10:43
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Apply the root test to the above power series. A careful examination gives option $(D)$.

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Using Cauchy root test we have:$$\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{3^{1+\frac{2}{n}}}|4x-12|\frac{1}{(n^2+1)^{\frac{1}{n}}}=\frac{1}{3}|4x-12|$$then for $|4x-12|<3$ the series converges and for $|4x-12|>3$ it diverges. Also for $|4x-12|=3$ we have:$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{1}{n^{2}+1}$$or$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{3^{n+2}}\frac{1}{n^{2}+1}$$which converges in both cases. Then the domain is $|4x-12|\le 3\to2.25\le x\le 3.75$

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  • $\begingroup$ Sir why you applied absolute convergence test? $\endgroup$ – Mohan Sharma Jan 24 '18 at 10:58
  • $\begingroup$ This is what defined by root test I took it form here:en.wikipedia.org/wiki/Root_test $\endgroup$ – Mostafa Ayaz Jan 24 '18 at 11:02
  • $\begingroup$ I have added my books image $\endgroup$ – Mohan Sharma Jan 24 '18 at 11:10
  • $\begingroup$ Then B is true according to my effort... $\endgroup$ – Mostafa Ayaz Jan 24 '18 at 11:12
  • $\begingroup$ Why you did not apply Cauchy's root test? $\endgroup$ – Mohan Sharma Jan 24 '18 at 11:13
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$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(9\right)}\frac{\left(4-\frac{4x}{3}\right)^{n}}{n^{2}+1}$$ according to Cauchy root test , we get $$|4-\frac{4x}{3}|\leqslant 1$$ so $$\frac{9}{4}\leqslant x\leqslant \frac{15}{4}$$

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