For which $x$ is $ \sum_{n=1}^{\infty} \frac{(x-b)^{n}}{na^{n}}$ (absolutely) convergent?

Question

Consider the power series: $$ \sum_{n=1}^{\infty} \frac{(x-b)^{n}}{na^{n}}$$ with a, b >0 :

a) for which x is this series absolute convergent,

b) for which x is this series conditionally convergent,

c) for which x is this series divergent.

So far this is what I have using the ratio test:

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(x-b)^{n+1}}{(n+1)a^{n+1}}}{\frac{(x-b)^n}{na^n}} = \frac{(x-b)na^n}{na^{n+1}+a^{n+1}} = \frac{(x-b)}{na+a^{n+1}}$$

I don't know how to solve the rest of the question, can someone help me with this?

Answer 1

What about the root test? One has that \begin{align*} \limsup_{n\rightarrow\infty}|a_{n}|^{1/n} = \limsup_{n\rightarrow\infty}\left|\frac{x-b}{n^{1/n}a}\right| = \left|\frac{x-b}{a}\right| \end{align*}

This is because \begin{align*} \lim_{n\rightarrow\infty}n^{1/n} = \lim_{n\rightarrow\infty}\exp\left(\frac{\ln(n)}{n}\right) = \exp\left(\lim_{n\rightarrow\infty}\frac{\ln(n)}{n}\right) = \exp(0) = 1 \end{align*}

Consequently, the given series converges when \begin{align*} \left|\frac{x-b}{a}\right| < 1 \Longleftrightarrow b - a < x < b + a \end{align*}

And it diverges when \begin{align*} \left|\frac{x-b}{a}\right| > 1 \Longleftrightarrow (x > b + a)\vee(x < b - a) \end{align*}

Now it remains to study two cases. More precisely, \begin{align*} \displaystyle\frac{x-b}{a} = 1 \Longrightarrow \sum_{n=1}^{\infty}\frac{(x-b)^{n}}{na^{n}} = \sum_{n=1}^{\infty}\frac{1}{n} \longrightarrow \infty \end{align*} and we do also have \begin{align*} \frac{x-b}{a} = -1 \Longrightarrow \sum_{n=1}^{\infty}\frac{(x-b)^{n}}{na^{n}} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} < \infty \end{align*} The last statement results from the application of the alternating test for series, also known as the Leibniz test.

Hopefully this helps.

Answer 2

Actually$$\left|\frac{a_{n+1}}{a_n}\right|=\frac n{n+1}\times\frac1{|a|}\times|x-b|$$and, since$$\lim_{n\to\infty}\frac n{n+1}\times\frac1{|a|}\times|x-b|=\frac{|x-b|}{|a|},$$your series converges absolutely if $|x-b|<a$ and diverges if $|x-b|>a$.

If $x=a+b$, then your series becomes$$\sum_{n=1}^\infty\frac1n,$$which diverges. And if $x=-a+b$, then your series becomes$$\sum_{n=1}^\infty\frac{(-1)^n}n,$$which converges, by the Leibniz test.

Answer 3

The radius of convergence $R=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$. In your case , for the given series to be convergent $$L=\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to\infty}|(x-b)|\frac{n}{a(n+1)}$$ The ratio test tells us that if $L<1$, the series converges, if $L>1$, the series divereges. $$L=\lim_{n\to\infty}\frac{n}{n+1}\frac{1}{a}|x-b|=\frac{1}{a}|x-b|$$ Now, $$\frac{1}{a}|x-b|<1\; \text{ series converges}$$ $$\frac{1}{a}|x-b|>1\; \text{ series diverges}$$

You may check what happens at the boundary points i.e. when $L=1$.