Why is the graph of $y=\displaystyle\lim_{k\rightarrow\infty}(\sum_{x=0}^k\frac{1}{2^x})$ discontinuous?

Question

I noticed some discontinuties in the graph of-

$$y=\displaystyle\lim_{k\rightarrow\infty}(\sum_{x=0}^k\frac{1}{2^x})$$

This is what is looks like on the graph-

Why is the graph discontinuous?I know that functions may be discontinuous if their value cannot be determined for a particular value (or it is not defined for that value or simply it does not exist).But here,the value of $y$ can be determined for all $x$.

Then why is the graph discontinuous?

Thanks for any help and response!!

Answer 1

Consider the function $$y_k=\sum_{x=0}^k{a^x}$$ It is just a geometric progression which makes $$y_k=\frac{1-a^{k+1}}{1-a}$$ If, as in you case $a=\frac 12$, then $y_k=2-2^{-k}=2-e^{-k\log(2)}$ which would be continuous if $k$ were continuous. Since this is not the case, you need a discrete plot of the following values $$\left( \begin{array}{cc} k & y_k \\ 0 & 1 \\ 1 & \frac{3}{2} \\ 2 & \frac{7}{4} \\ 3 & \frac{15}{8} \\ 4 & \frac{31}{16} \\ 5 & \frac{63}{32} \\ 6 & \frac{127}{64} \\ 7 & \frac{255}{128} \\ 8 & \frac{511}{256} \\ 9 & \frac{1023}{512} \\ 10 & \frac{2047}{1024} \end{array} \right)$$