What is the interval of convergence of #sum_1^oo [(3x)^n(x-2)^n]/(nx)
#?

Question

What is the interval of convergence of  #sum_1^oo [(3x)^n(x-2)^n]/(nx)
#?

Ezra Adams · Accepted Answer

#sum_1^oo [(3x)^n(x-2)^n]/(nx)=-(LogAbs[3 (x-2) x-1]]/x# and converges for #x ne 0#
#1/3 (3 - 2 sqrt[3]) < x < 1/3 (3 - sqrt[6])#
#1/3 (3 + sqrt[6]) < x < 1/3 (3 + 2 sqrt[3])# #sum_1^oo [(3x)^n(x-2)^n]/(nx) = 1/x sum_1^oo [(3x)^n(x-2)^n]/n# but #d/(dx)( ((3x)^n(x-2)^n)/n) = 6 (x - 1) ((3 x) (x - 2))^(k - 1)# or #d/(dx)(sum_1^oo [(3x)^n(x-2)^n]/n)=6(x-1)sum_1^{oo}((3 x) (x - 2))^(k - 1)# Considering now that #abs(3 x (x - 2))<1# #sum_1^{oo}((3 x) (x - 2))^(k - 1) = -1/(3 x (x - 2) - 1)# then #d/(dx)(sum_1^oo [(3x)^n(x-2)^n]/n) = -(6 (x - 1))/(3 x (x - 2) - 1)# Integrating again and dividing by #x# we obtain #sum_1^oo [(3x)^n(x-2)^n]/(nx) =-(LogAbs[3 (x-2) x-1]]/x #

Scarlett Allison · Answer

undefined To find the interval of convergence of the series $\sum_{n=1}^{\infty} \frac{(3x)^n(x-2)^n}{nx}$, we can use the ratio test.

Let $a_n = \frac{(3x)^n(x-2)^n}{nx}$. Then, the ratio test states that if $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$, the series converges absolutely.

We have:

$$
\begin{aligned}
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{\frac{(3x)^{n+1}(x-2)^{n+1}}{(n+1)x}}{\frac{(3x)^n(x-2)^n}{nx}} \right| \
&= \lim_{n \to \infty} \left| \frac{(3x)(x-2)(n+1)}{nx} \right| \
&= \lim_{n \to \infty} \left| \frac{3x(x-2)(n+1)}{x} \right| \
&= \lim_{n \to \infty} \left| 3(x-2)(n+1) \right| \
&= \lim_{n \to \infty} 3|n+1| \
&= 3\lim_{n \to \infty} |n+1| \
&= \infty
\end{aligned}
$$

Since the limit is greater than 1 for all values of $x$, the series diverges for all values of $x$.

What is the interval of convergence of #sum_1^oo [(3x)^n(x-2)^n]/(nx) #?