Can I replace $x$ with $t=2x+1$ to find the radius of convergence of $\sum_{n=1}^\infty\left(1+\frac{(-1)^n}n\right)^{n^2}\cdot\frac{(2x+1)^n}n$?

Question

I am requesting some help to spot where I did mistake since I spent hours reviewing my solution but wasn't successful spotting the mistake to correct it.

I need to find the radius of convergence for the following summation:

$$ f(x) = \sum_{n=1}^\infty \left(1+\frac{(-1)^n}{n}\right)^{n^2} \cdot \frac{(2x+1)^n}{n}$$ So, I defined a new function $f(t)$ that's exactly like $f(x)$ but replaced $2x+1$ with $t$ I calculated the radius of convergence for $f(t)$ and found that it's $1/e$

Since $t=2x+1$ thus $x=\frac{t-1}{2}$ I said that the radius of convergence for $f(x)-R_x$ is equal to $$\frac{R_{t}-1}{2}=\frac{\frac{1}{e}-1}{2}$$

But, I can confirm that my answer is wrong (through automatic checker).

Answer 1

Note that\begin{align}\limsup_n\sqrt[n]{\left|\left(1+\frac{(-1)^n}n\right)^{n^2}\frac{(2x+1)^n}n\right|}&=\limsup_n\left(1+\frac{(-1)^n}n\right)^n\frac{|2x+1|}{\sqrt[n]n}\\&=e|2x+1|\\&=2e\left|x+\frac12\right|.\end{align}Therefore, your series converges if $2e\left|x+\frac12\right|<1$ and diverges if $2e\left|x+\frac12\right|>1$. So, the radius of convergence is $\frac1{2e}$.

Answer 2

The series converges for

$$-\frac1e<t<\frac1e$$

or

$$-\frac1{2e}-\frac12<x<\frac1{2e}-\frac12.$$

Caution: the radius of convergence is the half-difference of theses bounds, not the upper bound.

Hence, $$\frac1{2e}.$$

Answer 3

Perhaps a little review of what is said about power series in your notes or textbook would help. There are also on-line resources such as https://tutorial.math.lamar.edu/classes/calcii/PowerSeries.aspx

In general, a power series is equivalent to $$\sum_{n = 0}^\infty c_n (x - a)^n.$$

The coefficient $c_n$ can be as complicated as you want as long as it doesn't depend on $x$ in any way.

If you have a power series in this form then you know it converges when $x=a$ because all terms after the first one come out to zero.

If the radius of convergence is $R$ then you know the series converges whenever $a - R < x < a + R$ and diverges whenever $x < a - R$ or $x > a + R.$

This is why one answer says to take half the difference between the two bounds of convergence. The two bounds are $a - R$ and $a + R$, the difference between them is $$ (a + R) - (a - R) = 2R, $$ and of course $R$ is half of $2R.$

In any case the radius is always centered at $a$ when the series is expressed in the form above. But you must have been misread whatever you were reading when you got the idea that the radius is centered at $1$ if the series is in $(x + 1).$ In order for the center to be $1$ you must have $a = 1$ in the formula above, and that means you have a series where each term has a factor $(x - 1)^n$, not $(x + 1)^n.$ The only way to fit a series with $(x + 1)^n$ into the form above is if $a = -1.$ If $a = -1$ then $(x - a)^n = (x + 1)^n$ and the radius of convergence of the series is centered at $-1.$

What if you have $(kx - a)^n$? Then the center generally isn't $a.$ You can factor out the $k$ like this:

$$ (kx - a)^n = k^n \left(x - \frac ak\right)^n. $$

In order to write such a series in the form $\sum_{n = 0}^\infty c_n (x - a)^n,$ The factor $k^n$ has to be part of the coefficient $c_n$, and the center is actually $\frac ak.$ But an easy way to find the center is just to set $kx - a = 0$ and solve for $x.$

In your series you have $(2x + 1)^n$ multiplied by some things that don't depend on $x.$ You can deal with this by factoring out the $2$ (actually $2^n$) to get a power series over $(x - a)$, if you can see what the value of $a$ must be. Or you can use the trick from the previous paragraph: just solve for $x$ in $2x + 1 = 0.$