Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2}  \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) \ \ \text{ none of } (\text{a})-(\text{c}).\end{equation*}
$$$$
I have done the following :
(a) It cannot be true since $2\cdot \left (2^{2022}-1\right )$ exceeds the sum of the series, which is a subseries of the infinite geometric series $\left \{\frac{1}{2^n}\right \}$ which converges to
$\frac{1}{1-\frac{1}{2}}=2$.
So we have that $2\cdot \left (2^{2022}-1\right )\gg2>S$.
(b) $\frac{2^{2022}-1}{2}=2^{2021}-\frac{1}{2}\gg2>S$
Therefore it cannot be true.
(c) We want to check if $S=\frac{2^{2022}-1}{2^{2021}}=2-\frac{1}{2^{2021}}$.
Let $\hat{S}$ be the sum of the infinite series : \begin{equation*}\hat{S}=1+\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2^n}=2 \text{ when } n\to \infty\end{equation*}
We therefore want to check if $S=\hat{S}-\frac{1}{2^{2021}}$ or $\hat{S}-S=\frac{1}{2^{2021}}$ ?
We know that $\hat{S}-S=\frac{1}{2^{2023}}+\frac{1}{2^{2024}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty \ \ \ (\star)$
We can rewrite $\frac{1}{2^{2021}}=\frac{2}{2^{2022}}=\frac{1}{2^{2022}}+\frac{1}{2^{2022}}=\frac{1}{2^{2022}}+\frac{2}{2^{2023}} \Rightarrow$ etc $\Rightarrow \frac{1}{2^{2021}}=\frac{1}{2^{2022}}+\frac{1}{2^{2023}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty$
Comparison with (⋆) gives that $\frac{1}{2^{2021}}>\hat{S}-S \Rightarrow $ c) cannot be true.
So (d) must be the correct answer.
$$$$
Is everything correct ?
 A: $\sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}$ $($ see here$) $
$\begin{align}S &=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} \\\end{align}$
Put $r=\frac{1}{2}, n=2022$.
Hence $S=2 (1-\frac{1}{2^{2023}})=\frac{2^{2023}-1}{2^{2022}}$
A: If $S = 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2021}} + \frac{1}{2^{2022}}$, then just add $\color{blue}{\frac{1}{2^{2022}}}$ to both sides to get the result:
\begin{align}
S + \color{blue}{\frac{1}{2^{2022}}} &= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \frac{1}{2^{2021}} + \underbrace{\frac{1}{2^{2022}} + \color{blue}{\frac{1}{2^{2022}}}}\\
&= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \underbrace{\frac{1}{2^{2021}} + \ \ \ \ \ \ \ \frac{1}{2^{2021}}}\\
&= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2^{2020}}\\
&= \cdots\\
&= 1 + 1\\
&= 2\\
\ \\
\implies S &= 2 - \frac{1}{2^{2022}}
\end{align}
A: I think you are correct, although your answer for $(c)$ is unnecessarily long-winded. You would benefit from using the formula for a finite sum of geometric series, rather than using the infinite sum everywhere. It is:
$$ \sum_{k=1}^{n} ar^k = a\left( \frac{1-r^{n+1}}{1-r} \right).$$
where $a$ is the first term and $r$ is the common ratio.
Therefore:
$$ S = 1\left( \frac{1-\left(\frac{1}{2}\right)^{2023}}{1-\frac{1}{2}} \right)  = 2 - \left(\frac{1}{2}\right)^{2022}. $$
We didn't need to simplify like that for the purposes of this question, but it is the form that you can most easily see what $S$ is equal to - for future use. Anyway, so answer $(c)$ what we can do is:
$$ S = \frac{2^{2022}\left(1-\left(\frac{1}{2}\right)^{2023}\right)}{2^{2022}\left(\frac{1}{2}\right)} = \frac{2^{2022}-\frac{1}{2}}{2^{2021}} \neq\frac{2^{2022}-1}{2^{2021}}.$$
