MHB What is the answer after convergence in TI-Nspire CX CAS?

[karush]

$\text{Evaluate answer from }\textit{ TI-Nspire CX CAS}$
\begin{align*}
\displaystyle
S_k&=\sum_{k=1}^{\infty}
\left[\frac{(-2)}{9^{k+1}}\right]
=\frac{-2}{99} \\
\end{align*}
ok wasn't sure what weapon of choice to use
$\tiny{206.10.3.75}$

 

[MarkFL]

$$S=\sum_{k=1}^{\infty}\left(\frac{-2}{9^{k+1}}\right)=-\frac{2}{81}\sum_{k=1}^{\infty}\left(\left(\frac{1}{9}\right)^{k-1}\right)=-\frac{2}{81}\cdot\frac{1}{1-\dfrac{1}{9}}=-\frac{2}{81}\cdot\frac{9}{8}=-\frac{1}{36}$$

 

[karush]

$\text{Evaluate }$
$$\displaystyle
S_k=\sum_{k=1}^{\infty}
\left[\frac{(-2)}{9^{k+1}}\right]
=\frac{-2}{99} $$
$\text{geometric series}$
$$\sum_{k=1}^{\infty}ar^{k-1}=\frac{a}{(1-r)},
\ \ \left| r \right|< 1$$
$$\displaystyle S_k=-\frac{2}{81}\sum_{k=1}^{\infty}
\left(\frac{1}{9}\right)^{k-1}
=\frac{1}{1-\dfrac{1}{9}}
=-\frac{2}{81}\cdot\frac{9}{8}=-\frac{1}{36}$$
$\tiny{206.10.3.71}$

 

[karush]

$\text{Evaluate }$
$$\displaystyle
S_k=\sum_{k=1}^{\infty}
\left[\frac{(-2)}{9^{k+1}}\right]
=\frac{-1}{3} $$
$\text{geometric series}$
$$\sum_{k=1}^{\infty}ar^{k-1}=\frac{a}{(1-r)},
\ \ \left| r \right|< 1$$
$$\displaystyle S_k=-\frac{2}{81}\sum_{k=1}^{\infty}
\left(\frac{1}{9}\right)^{k-1}
=\frac{1}{1-\dfrac{1}{9}}
=-\frac{2}{81}\cdot\frac{9}{8}=-\frac{1}{36}$$
$\tiny{206.10.3.75}$