MHB Ratio Test for Sum $\tiny{206.10.5.84}$

[karush]

$\tiny{206.10.5.84}$
\begin{align*}
\displaystyle
S_{84}&=\sum_{k=1}^{\infty}
\frac{(4x)^k}{5k}\\

\end{align*}
$\textsf{ ratio test}$
$$\frac{a_{n+1}}{a_n}
=\frac{ \frac{(4x)^{k+1}}{5(k+1)}}{ \frac{(4x)^k}{5k}}
=\frac{4xk}{k+1} $$
$\textsf{W|A says this converges at $4|x|<1 $ so how??}$

 

[Greg]

The ratio test says a sum converges absolutely if

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

We have

$$\lim_{k\to\infty}\left|\frac{4xk}{k+1}\right|=4|x|$$

so the series converges absolutely if $4|x|<1$.

 

[Greg]

$$\sum_{k=1}^\infty\frac{(4x)^k}{5k}=\frac{4x}{5}+\frac{16x^2}{10}+\frac{64x^3}{15}+\frac{256x^4}{20}+\frac{1024x^5}{25}+\cdots$$

$$\frac{\text d}{\text{d}x}\left(\sum_{k=1}^\infty\frac{(4x)^k}{5k}\right)=\frac45+\frac{16x}{5}+\frac{64x^2}{5}+\frac{256x^3}{5}+\frac{1024x^4}{5}+\cdots$$

$$S=\int\frac45\left(\sum_{k=1}^\infty(4x)^{k-1}\right)\,\text{d}x,\quad\text{ which converges for }|x|<\frac14$$

$$S=\frac45\int\frac{1}{1-4x}\,\text{d}x=\frac45\cdot-\frac14\log(1-4x)+C=-\frac15\log(1-4x),\quad|x|<\frac14;\,C=0$$

 

[karush]

this has been a difficult section so the help here has been very appreciated..