Limit $ \sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j} \left(-\frac{1}{3}\right)^j \right) $ I have to find the limit of the following series:
$$ \sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j} \left(-\frac{1}{3}\right)^j \right) $$
I don't even know how to approach this... Any help would be very appreciated
 A: Using the binomial formula and the geometric series formula:
$$\sum_{k=0}^{\infty}\left(\sum_{j=0}^{k}{k\choose j}\left(-\frac13\right)^j\right)=\sum_{k=0}^{\infty}\left(1-\frac13\right)^k=\lim_{k\to\infty}\frac{(2/3)^{k+1}-1}{(2/3)-1}=\frac1{1-(2/3)}=3$$
A: Hint: The inner sum is just$$\sum_{j=0}^k\binom kj\left(-\frac13\right)^j=\left(1-\frac13\right)^k.$$
A: Note that according to the Newton's binomial theorem, $$ \sum_{j=0}^∞ \binom{k}{j} (x)^j =(1+x)^k$$
For $x= (-1/3)$ we get $$ \sum_{j=0}^∞ \binom{k}{j} (x)^j =(1+x)^k= (2/3)^k$$
Thus we have $$\sum_{k=0}^∞ \left( \sum_{j=0}^∞ \binom{k}{j} \left(-\frac{1}{3}\right)^j \right)= \frac {1}{1-2/3} =3$$
A: Since,
by the binomial theoram,
$ \sum_{j=0}^k \binom{k}{j}x^j 
=(1+x)^k
$,
$\begin{array}\\
\sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j}x^j \right)
&=\sum_{k=0}^∞ (1+x)^k\\
&=\dfrac{1}{1-(1+x)}
\qquad\text{geometric series with ratio }1+x\\
&=\dfrac{-1}{x}\\
\end{array}
$
Putting $x=-\frac13$,
this gives
$\dfrac{-1}{-\frac13}
=3$.
Note that the sum
does not converge
if $x > 0$,
or else you would
get the nonsensical result
(if, say,
$x=\frac12$),
$\sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j}(1/2)^j \right)
=-2$.
