How to Use Absolute Convergence Test on This Series?

[karush]

$$\sum_{k=1}^{\infty}\left(\frac{k}{k+1}\right)^{2k}$$
$\textsf{how do you use absolute converge test on this?}$

 

[MarkFL]

I think what I would do here is compute:

\(\displaystyle L=\lim_{k\to\infty}\left(\left(\frac{k}{k+1}\right)^{2k}\right)\)

If $L\ne0$, then the series must diverge. Bear in mind though, that this test is inconclusive if the limit of the summand is zero.

 

[karush]

$\textsf{Find the radius / interval of convergence }\\$
\begin{align}
\displaystyle f(x)&=2k(x-1)^k
\end{align}
$\textsf{thot would ask another questions here since new stuff?}$

 

[MarkFL]

$\textsf{Find the radius / interval of convergence }\\$
\begin{align}
\displaystyle f(x)&=2k(x-1)^k
\end{align}
$\textsf{thot would ask another questions here since new stuff?}$

If we are given:

\(\displaystyle f(x)=\sum_{k=0}^{\infty}\left(2k(x-1)^k\right)\)

Then we compute the radius of convergence as follow:

\(\displaystyle |x-1|<\lim_{k\to\infty}\left|\frac{2k}{2(k+1)}\right|=1\)

 

[karush]

If we are given:

\(\displaystyle f(x)=\sum_{k=0}^{\infty}\left(2k(x-1)^k\right)\)

Then we compute the radius of convergence as follow:

\(\displaystyle |x-1|<\lim_{k\to\infty}\left|\frac{2k}{2(k+1)}\right|=1\)

$\text{what happened to the exponent $k$ or that a concern?}$

 

[MarkFL]

We are only concerned with the coefficient of the exponential term. In fact we can even remove any constant factors in that coefficient.