Secret age

This is actually a very simple problem in disguise. It can be rewritten as: $\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ 2{ \left( \frac { 3966 }{ 3965 } \right) }^{ n } } }$ This makes the problem a whole lot easier. Let ( $S$ ) be the year she was born, and ( $x$ ) be $\left( \frac { 3966 }{ 3965 } \right)$ . $s=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ 2{ \left( x \right) }^{ n } } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2x } +\frac { 1 }{ 2{ x }^{ 2 } } ...$ $\frac { s }{ x } =\frac { 1 }{ 2x } +\frac { 1 }{ 2{ x }^{ 2 } } +\frac { 1 }{ 2{ x }^{ 3 } } ...$ $s-\frac { s }{ x } =s\left( 1-\frac { 1 }{ x } \right) =\frac { 1 }{ 2 }$ $s=\frac { \left( \frac { 1 }{ 2 } \right) }{ \left( 1-\frac { 1 }{ x } \right) } =\frac { 1 }{ 2\left( 1-\frac { 1 }{ x } \right) }$ Substitute $x=\left( \frac { 3966 }{ 3965 } \right)$ : $s=\frac { 1 }{ 2\left( 1-\frac { 3965 }{ 3966 } \right) } =\frac { 3966 }{ 2 } =1983$ Since it is the year $2014$ , her age is $2014-1983=\boxed{31}$

By the way, in case my Maths teacher decides to get a brilliant account and sees this and tries to kill me, $31$ isn't her real age.

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>>>from sympy import summation
>>>i = symbols('i', integer=True)
>>>summation(((3966.0)/(3965))**-(i+log(2.0)/log(3966.0/3965)), (i, 0, oo))
>>>1982.99999999998
>>>2014-1983
>>>31

Python will do the job!:)

Assume the given summand as y and let x = 3966/3965.

Thus, x^{-(n + log2(base x))} = y. Take log on both sides and solve. You get y = 0.5x^-n.

Now, the whole thing turns out to be an infinite GP (geometric progression) with common ratio 1/x. Whence, we get the result as x/2(x-1).

Substitute to get the result as 1983.

Hence age = 31 yrs

you can simply transform this sum into a integral, because every Riemann's sum with infinite terms can be written as a integral, like this: $\lim_{k\rightarrow \infty} \int_0^k \! \frac{3966}{3965}^{-(n+\log_{\frac{3966}{3965}} 2)} \mathrm{d}n.$

then:

$\int_0^{\infty} \! \frac{3966}{3965}^{-(n+\log_{\frac{3966}{3965}} 2)} \mathrm{d}n = 1982.7$

So, $2014-1982.7=31.3$