Limit $ \sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j} \left(-\frac{1}{3}\right)^j \right) $

Question

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

Answer 1

Using the binomial formula and the geometric series formula: $$\sum_{k=0}^{\infty}\left(\sum_{j=0}^{k}{k\choose j}\left(-\frac13\right)^j\right)=\sum_{k=0}^{\infty}\left(1-\frac13\right)^k=\lim_{k\to\infty}\frac{(2/3)^{k+1}-1}{(2/3)-1}=\frac1{1-(2/3)}=3$$

Answer 2

Hint: The inner sum is just$$\sum_{j=0}^k\binom kj\left(-\frac13\right)^j=\left(1-\frac13\right)^k.$$

Answer 3

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 $$\sum_{k=0}^∞ \left( \sum_{j=0}^∞ \binom{k}{j} \left(-\frac{1}{3}\right)^j \right)= \frac {1}{1-2/3} =3$$

Answer 4

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, $x=\frac12$), $\sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j}(1/2)^j \right) =-2$.