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 can i split $\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}$ into 2 sums?
2
$\begingroup$
sequences-and-series
summation
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0How about $2(\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}) - \sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k}*(\frac{4}{7})^{n+k}$? – 2017-01-29
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0I dont really think that this would help me out with finding the value of the summation or whether it converges. – 2017-01-29
1 Answers
1
$$\sum_{n=0}^\infty\sum_{k=0}^n\binom nk\left(\dfrac47\right)^{n+k}$$
$$=\sum_{n=0}^\infty\left(\dfrac47\right)^n\sum_{k=0}^n\binom nk\left(\dfrac47\right)^k$$
$$=\sum_{n=0}^\infty\left(\dfrac47\right)^n\left(1+\dfrac47\right)^n\text{ using }(1+x)^n=\sum_{k=0}^n\binom nk x^k$$
$$=\sum_{n=0}^\infty\left(\dfrac{4\cdot11}{7^2}\right)^n$$
$$=\dfrac1{1-\dfrac{4\cdot11}{7^2}}$$ as $\left|\dfrac{4\cdot11}{7^2}\right|<1$