Birthday Problem!

$\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 6n }{ { 4 }^{ n } } }=6 \displaystyle \sum _{ n=1 }^{ \infty }{ \frac {n }{ { 4 }^{ n } } } \\ 6\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+\ldots \right) \\ \text{Let }S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+\ldots \\ \frac{S}{4}=\frac{1}{4^2}+\frac{2}{4^3}+\frac{3}{4^4}+\frac{4}{4^5}+\ldots \\ S-\frac{S}{4} =\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4} \\ \frac{3S}{4}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3} \\ S=\frac{4}{3}\times\frac{1}{3}=\frac{4}{9} \\ \therefore 6S=\frac{8}{3}=\boxed{2.66}$

I made this formula for k>1 $\sum_{n=1}^{\infty}{\frac{n}{k^n}}=\frac{k}{(k-1)^2},$ then $6\sum_{n=1}^{\infty}{\frac{n}{4^n}=6\frac{4}{3^2}=\frac{8}{3}=2.\overline{6}}$

Note that $\frac{1}{1-x}=1+x+x^2+x^3+\ldots +x^n +\ldots , \, \, \, \text{where}\, |x|<1$ Differentiate it and multiple both sides with $x$ , we obtain $\frac{x}{(1-x)^2}=x+2x^2+3x^3+\ldots + nx^n + \ldots, \, \, \, \text{where}\, |x|<1$

Note that $\sum _{ n=1 }^{ \infty }{ \frac { 6n }{ { 4 }^{ n } } }=6\left( x+2x^2+3x^3+\ldots + nx^n + \ldots,\right)\, \, \, \text{where}\, x=\frac{1}{4}$

Hence, $\sum _{ n=1 }^{ \infty }{ \frac { 6n }{ { 4 }^{ n } } }=6\times \frac{\frac{1}{4}}{(1-\frac{1}{4})^2}=\frac{8}{3}$

The sum is equal to $6\cdot \left(\frac 1{4^1} + \frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4} + \frac1{4^5} + \cdots \\ +\frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4} + \frac 1{4^5} + \frac1{4^6} + \cdots \\ +\frac 1{4^3} + \frac 1{4^4} + \frac 1{4^5} + \frac 1{4^6} + \frac1{4^7} + \cdots \\ + \cdots\right) \\ = 24 \cdot \left\{ \frac 1{4^1} \cdot \left(\frac 1{4^1} + \frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4} + \frac1{4^5} + \cdots \right) \\ + \frac 1{4^2} \cdot \left(\frac 1{4^1} + \frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4} + \frac1{4^5} + \cdots \right) \\ + \frac 1{4^3} \cdot \left(\frac 1{4^1} + \frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4} + \frac1{4^5} + \cdots \right) \\ + \cdots\right\} \\ = 24 \cdot \left(\frac 1{4^1} + \frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4}+ \cdots\right)\cdot \left(\frac 1{4^1} + \frac 1{4^2} + \frac 1{4^3} + \frac 1{4^4}+ \cdots\right) \\ = 24 \cdot \left(\sum_{n=1}^\infty (\tfrac14)^n\right)^2 = 24\cdot (\tfrac13)^2 = \tfrac83 \approx \boxed{2.67}.$