How can i split $\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}$ into 2 sums?

Hi i would like to split $\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}$

into two summations. Is this even possible ?

• How about $2(\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}) - \sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}$? – OFRBG Jan 29 '17 at 15:00
• I dont really think that this would help me out with finding the value of the summation or whether it converges. – asddf Jan 29 '17 at 15:02

$$\sum_{n=0}^\infty\sum_{k=0}^n\binom nk\left(\dfrac47\right)^{n+k}$$
$$=\sum_{n=0}^\infty\left(\dfrac47\right)^n\sum_{k=0}^n\binom nk\left(\dfrac47\right)^k$$
$$=\sum_{n=0}^\infty\left(\dfrac47\right)^n\left(1+\dfrac47\right)^n\text{ using }(1+x)^n=\sum_{k=0}^n\binom nk x^k$$
$$=\sum_{n=0}^\infty\left(\dfrac{4\cdot11}{7^2}\right)^n$$
$$=\dfrac1{1-\dfrac{4\cdot11}{7^2}}$$ as $\left|\dfrac{4\cdot11}{7^2}\right|<1$