Question

Consider the series

Answer 1

Answer:

Step-by-step explanation:

Recall the ratio test. Given a series $\sum_{n=1}^{\infty}a_n$ if

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right|$

Then, the series is absolutely convergent.

We will use this to the given series $\sum_{n=1}^{\infty} \frac{(2 x)^n}{n}$, where $a_n = \frac{(2 x)^n}{n}$. Then, we want to find the values for which the series converges.

So

$\lim_{n\to \infty} \left|\frac{(2x)^{n+1}}{n+1}\cdot \frac{n}{(2x)^n}\right|$, which gives us that

$|2x|\cdot\lim_{n\to \infty} \frac{n}{n+1}$

We have that $\lim_{n\to \infty} \frac{n}{n+1}=1$. Then, we have that

$|2x|$,

which implies that |x|<1/2. So for $x \in (-1/2,1/2)$ the series converges absolutely.

We will replace x by the endpoints to check convergence.

Case 1, x=1/2:

In this case we have the following series:

$\sum_{n=1}^{\infty} \frac{1}{n}$ which is the harmonic series, which is know to diverge.

Case 2, x=-1/2:

In this case we have the following series:

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$

This is an alternating series with $b_n = \frac{1}{n}$. Recall the alternating series test. If we have the following

$\sum_{n=1}^\infty (-1)^nb_n$ and$b_n$ meets the following criteria : bn is positive, bn is a decreasing sequence and it tends to zero as n tends to infinity, then the series converge.

Note that in this case, $b_n = \frac{1}{n}$ si always positive, its' limit is zero as n tends to infinity and it is decreasing, hence the series converge.

So, the final interval of convergence is

$[\frac{-1}{2}, \frac{1}{2})$

Answer 2

Answer:

(-\infty,-1/2) U (1/2,+\infty)

Step-by-step explanation:

You have the following series:

$\sum_{n=1}^{\infty} \frac{(2 x)^n}{n}$

You calculate the radius of convergence by using the formula:

$R= \lim_{n \to \infty} |\frac{a(x)_n}{a(x)_{n+1}}|= \lim_{n \to \infty} |\frac{\frac{(2x)^n}{n}}{\frac{(2x)^{n+1}}{n+1}}|\\\\=\lim_{n \to \infty} |\frac{\frac{(2x)^n}{n}}{\frac{(2x)^n(2x)}{n+1}}|=\lim_{n \to \infty}|\frac{n+1}{2xn}|=|\frac{1}{2x}|\lim_{n \to \infty}|1+\frac{1}{n}|=|\frac{1}{2x}|$

The radius of convergence is R=1/2x.

Hence, the interval of convergence is

|2x| < 1

|x| < 1/2

By evaluating in the extrems of the interval:

$\sum_{n=1}^{\infty} \frac{(2 (\frac{1}{2}))^n}{n}=\sum_{n=1}^{\infty} \frac{(1)^n}{n}=0\\\\\sum_{n=1}^{\infty} \frac{(2 (-\frac{1}{2}))^n}{n}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$

for x=-1/2 we obtain an Alternating Harmonic Series, for x=1/2 we obtain the divergent harmonic series. Thus the interval is:

(-\infty,-1/2) U [1/2,+\infty)