find the domain of convergence of the given series $\sum_{n=1}^{\infty}\frac{ n.4^n }{3^n}x^n(1-x)^n$

Question

find the domain of convergence of the given series

$$\sum_{n=1}^{\infty}\frac{ n.4^n }{3^n}x^n(1-x)^n$$

My attempt : $\frac{ n.4^n }{3^n}x^n(1-x)^n = n.(\frac{4}{3}x - \frac{4}{3}x^2)^n$ now $$-1 <|n.(\frac{4}{3}x - \frac{4}{3}x^2)^n| < 1$$

= $-1< n\frac{4}{3}(x- x^2)< 1$

=$-\frac{3}{4}< n(x- x^2)< \frac{3}{4}$

Now im not able to proceed Further pliz help me

thanks u

Answer 1

Note that$$\lim_{n\to\infty}\sqrt[n]{\left\lvert\frac{n4^n}{3^n}x^n(1-x)^n\right\rvert}=\lim_{n\to\infty}\sqrt[n]n\frac43\bigl\lvert x(1-x)\bigr\rvert=\frac43\bigl\lvert x(1-x)\bigr\rvert.$$Therefore, by the root test, the series converges if $\bigl\lvert x(1-x)\bigr\rvert<\frac34$ and diverges if $\bigl\lvert x(1-x)\bigr\rvert>\frac34$. Actually, it also diverges when $\bigl\lvert x(1-x)\bigr\rvert=\frac34$, because then we dont't have$$\lim_{n\to\infty}\left\lvert\frac{n4^n}{3^n}x^n(1-x)^n\right\rvert=0.$$So, the series converges if and only if $\bigl\lvert x(1-x)\bigr\rvert<\frac34$.

Answer 2

At the end you made a mistake in using the ratio test. It should be $|\frac{n}{n-1}(x-x^2)|\lt \frac{3}{4}$. Since $\frac{n}{n-1}\to 1$, you need to test $|x-x^2|\lt \frac{3}{4}$. The solution is $-\frac{1}{2}\lt x \lt \frac{3}{2}$.