Is there a value of r for which $\sum_{k=0}^{\infty}r^k = \frac{1}{4}$? Is there a value of r for which $\sum_{k=0}^{\infty}r^k = \frac{1}{4}$?
If I just plug the values into the formula I get:
$\frac{1}{1-r}$ = $\frac{1}{4}$
$r=-3$
But something tells me this can't be right? I would love any feedback!
 A: Here is a rather informal, but hopefully intuitively approach, as to why that sum can never be $1/4$. You may recognize your series as a standard Geometric series for which , in order to converge, we must have $-1<r<1$. (r=0 conveniently excluded).Now if you pick $r=-999/1000$ you find a sum of $1000/1999$ ,which is just a tad above $1/2$. On the other end, if you pick $r=999/1000$, then the sum would be $1000$. You can try this out by choosing even closer values to $-1$ and $1$ on interval $(-1,1)$. Seems like that $1/2$ is a lower boundary (and there is no upper boundary). If you use a graphing tool (Desmos, TI83,84, etc) you can confirm that by examining the graph of $y=\frac{1}{1-x}$. In particular, check out that asymptote $x=1$.The desired output of $1/4$ is not in the graph's range on domain $(-1,1)$ In other words: There is a whole spectrum of numbers that cannot be had as a standard geometric sum
A: The sum can never be less than 1
for $r > 0$
because the first term is 1
and the other terms are positive.
If
$-1 < r < 0$,
the sum is
$\frac1{1-r}$.
If this is $\frac14$,
then,
as OP stated,
$r = -3$
which is outside the range
of convergence.
More generally,
if
$-1 < r < 0$,
then
$1 < 1-r < 2$
so
$\frac12 <
\frac1{1-r}
< 1$.
Therefore the sum is always
at least $\frac12$.
A: Let $f_n(r)=\sum_{k=0}^nr^k$. If $r=1$ then $f_n(r)=(n+1)$, which does not converge as $n\to \infty$.
If $r\ne 1$ then $f_n(r)=\frac {1-r^{n+1}}{1-r}.\;$ For $r\ne 1$ this does not converge as $n\to \infty$ unless $|r|<1.$
For $|r|<1$ we have $\lim_{n\to \infty}f_n(r)=\frac {1}{1-r}.$  Now $$\frac {1}{1-r}=\frac {1}{4}\iff r=-3\implies |r|>1,$$ but $\lim_{n\to \infty}f_n(r)$ does not exist when $|r|>1.$
So there is no value of $r$ for which $\sum_{k=0}^{\infty}r^k=\frac {1}{4}.$ 
