Question

Question regarding radius of converges for the series.
Help understand how to find ak.

Answer 1

The series is given by \(\sum a_k x^k\), so
\[a_k=\left(-\frac{1}{2}\right)^k\cdot\frac{1}{k}\]

Answer 2

why did they get 1/(k2^k) though?

Answer 3

They take the absolute value of the ratio \(\dfrac{a_k}{a_{k+1}}\), so the negative powers are canceled out.

Answer 4

still dont fully understand

Answer 5

Consider the factor of \(\left(-\dfrac{1}{2}\right)^k\):
\[\left|\left(-\frac{1}{2}\right)^k\right|=\left|(-1)^k\left(\frac{1}{2}\right)^k\right|=\left|(-1)^k\right|\left|\left(\frac{1}{2}\right)^k\right|=1\cdot\frac{1}{2^k}\]
\((-1)^k=\pm1\) for integer exponents \(k\). Regardless of what \(k\) you plug in, the absolute value will always take on the positive 1.

Answer 6

haha oh I see now. Thank you so much for that clarification! :)

Answer 7

yw

Answer 8

how did they get the lim part?

Answer 9

esp 2nd step

Answer 10

To conduct the ratio test, you have to take a limit of the ratio of two successive terms \(a_k\) and \(a_{k+1}\).

Think back to the convergence conditions of a geometric series. If the magnitude of the ratio \(r\) is less than 1, then the series converges (i.e. if \(|r|<1\)).

If you can show that the series behaves like a geometric series and you can establish that the ratio of successive terms is less than 1, then series must converges.

With power series like this one, the limit will contain a power of \(|x|\).

Consider a general power series \(\sum a_kx^k\). By the ratio test,
\[\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=|x|\lim_{k\to\infty}(\cdots)=|x|L\]
provided that the limit exists. For the series to converge, you must have that \(|x|L<1\), which means the series converges for \(|x|<\dfrac{1}{L}\).

Answer 11

If you were wondering more about the specific computation:
\[\begin{align*}\lim_{k\to\infty}\underbrace{\left|\left(-\frac{1}{2}\right)^{k+1}\frac{|x|^{k+1}}{k+1}\right|}_{\huge a_{k+1}}\underbrace{\left|\left(-\frac{1}{2}\right)^{-k}\frac{k}{|x|^k}\right|}_{\huge \frac{1}{a_k}}&=\lim_{k\to\infty}\left(\frac{1}{2}\right)^{k+1}\frac{|x|^{k+1}}{k+1}\left(\frac{1}{2}\right)^{-k}\frac{k}{|x|^k}\\
&=\lim_{k\to\infty}\frac{1}{2}\frac{|x|k}{k+1}\\
&=\frac{|x|}{2}\lim_{k\to\infty}\frac{k}{k+1}\\
&=\frac{|x|}{2}\cdot1
\end{align*}\]
Convergence occurs when the limit is less then 1, which means when
\[\frac{|x|}{2}<1~~\iff~~|x|<2\]
So 2 is the radius.

Answer 12

One thing to mention: where I carried out the entirety of the ratio test, the provided solution uses a shortcut. If \(R\) is the radius, then
\[R=\frac{1}{\displaystyle\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|}=\lim_{k\to\infty}\left|\frac{a_k}{a_{k+1}}\right|\]