Are you smarter than me? 4

$\sum _{ n=1 }^{ \infty }{ \frac { n+9 }{ { (2 }^{ n+9 })(n)(n+8) } }$

= $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2 }^{ n+9 }) } } \times \frac { n+9 }{ n(n+8) }$

= $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2 }^{ n+9 }) } } \left( \frac { 9 }{ 8(n) } +\frac { 1 }{ 8(n+8) } \right)$

= $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2 }^{ n+12 }) } \left( \frac { 9 }{ n } \right) } \quad +\quad \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2 }^{ n+12 }) } \left( \frac { 1 }{ n+8 } \right) }$

= $\frac { 9 }{ { 2 }^{ 12 } } \left( \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { 2 }^{ n }(n) } } \right) \quad +\quad \frac { 1 }{ { 2 }^{ 12 } } \left( \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { 2 }^{ n }(n+8) } } \right)$

Now, $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { 2 }^{ n }(n) } } = ln 2$

let $x = n+ 8$ in the 2nd series,

= $\frac { 9 }{ { 2 }^{ 12 } } \left( ln\quad 2 \right) \quad +\quad \frac { 1 }{ { 2 }^{ 12 } } \left( \sum _{ x=9 }^{ \infty }{ \frac { 1 }{ { 2 }^{ x-8 }(x) } } \right)$

= $\frac { 9 }{ { 2 }^{ 12 } } \left( ln\quad 2 \right) \quad +\quad \frac { { 2 }^{ 8 } }{ { 2 }^{ 12 } } \left( \sum _{ x=9 }^{ \infty }{ \frac { 1 }{ { 2 }^{ x }(x) } } \right)$

= $\frac { 9 }{ { 2 }^{ 12 } } \left( ln\quad 2 \right) \quad +\quad \frac { { 2 }^{ 8 } }{ { 2 }^{ 12 } } \left( ln\quad 2\quad -\quad \sum _{ n=1 }^{ 8 }{ \frac { 1 }{ { 2 }^{ n }(n) } } \right)$

Now, computing all that, we get the value as -

$\approx \boxed{0.00149}$

Eight lines of Python will work just fine.

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from fractions import Fraction as frac
total = 0
i = 1
infinity = 50
while i <= infinity:
    total += frac(i+9,(2**(i+9))*i*(i+8))
    i += 1
print float(total)