Ghany's sum

Denote the numerator sum as $S$ and denominator as $T$ . So,

$\begin{aligned} S &= \dfrac{1} {1^4} + \dfrac{1} {3^4} + \dfrac{1} {5^4} + \ldots \\ T &= \dfrac{1} {2^4} + \dfrac{1} {4^4} + \dfrac{1} {6^4} + \ldots \\ &= \dfrac{1} {2^4} \left( 1 + \dfrac{1} {2^4} + \dfrac{1} {3^4} + \ldots \right) \\ &= \dfrac{1} {2^4} \left( T + S \right) \end{aligned}$ .

Solving for $\dfrac S T$ gives the answer: $15$ .

Let $Q= \sum_{n=1}^{\infty} \dfrac{1}{(2n)^4}$

Let $P=\sum_{n=1}^{\infty} \dfrac{1}{(2n-1)^4}$

Observe that $(P+Q) \cdot \dfrac{1}{2^4}= Q$

which gives us, $P+Q=16.Q$

So, $\dfrac{P}{Q}=15$

$\frac{\sum \limits_{n=1}^{\infty} (2n-1)^{-4}}{\sum \limits_{n=1}^{\infty} (2n)^{-4}}$

$= \frac{\sum \limits_{n=1}^{\infty} (2n-1)^{-4} +\sum \limits_{n=1}^{\infty} (2n)^{-4}-\sum \limits_{n=1}^{\infty} (2n)^{-4}}{\sum \limits_{n=1}^{\infty} (2n)^{-4}}$ ( On Adding and subtracting the denominator in the numerator )

$=\frac{\sum \limits_{n=1}^{\infty} (n)^{-4} +-\sum \limits_{n=1}^{\infty} (2n)^{-4}}{\sum \limits_{n=1}^{\infty} (2n)^{-4}}$

$= \frac{\sum \limits_{n=1}^{\infty} (n)^{-4}}{\sum \limits_{n=1}^{\infty} (2n)^{-4}} -1$

$= \frac{\sum \limits_{n=1}^{\infty} (n)^{-4}}{2^{-4}\sum \limits_{n=1}^{\infty} (n)^{-4}} -1$

$= \frac{1}{2^{-4}} -1$

$=2^{4} -1 = 15$

First, let $\sum_{n=1}^\infty (2n-1)^{-4} = x$ , and $\sum_{n=1}^\infty (2n)^{-4} = y$ .

Then, we can say for $y$ that

$y = \sum_{n=1}^\infty \frac {1} {(2n)^{4}} = \frac {1} {16} \sum_{n=1}^\infty \frac {1} {n^{4}}$ .

Then, since

$\sum_{n=1}^\infty \frac {1} {n^{4}} = x + y$ ,

we can also say that

$x=\sum_{n=1}^\infty \frac {1} {n^{4}} - y$ .

Then, the fraction of the question can be changed into

$\frac {x} {y} = \frac {\sum_{n=1}^\infty \frac {1} {n^{4}} - y} {\frac {1} {16} \sum_{n=1}^\infty \frac {1} {n^{4}}}$ .

Then, we can simplify this equation into

$\frac {x} {y} = \frac {\sum_{n=1}^\infty \frac {1} {n^{4}}} {\frac {1} {16} \sum_{n=1}^\infty \frac {1} {n^{4}}} -1$ .

Then, the answer comes out to be

$16 - 1 = \boxed {15}$ .

$\sum_{n=1}^n$ {2n-1} = $\sum_{n=1}^n - \sum_{n=1}^{n/2}$ 2n

(Notice that, the limits for the second term (i.e, for $\sum_{n=1}^{n/2}$ 2n) goes up till n/2)

Here the number of terms is infinite and thus, half the number will also be infinite.

Therefore we can write

$\sum_{n=1}^{\infty} \ (2n-1)^{-4}$ = $\sum_{n=1}^{\infty} \ (n)^{-4} - \sum_{n=1}^{\infty} \ (2n)^{-4}$

$=\sum_{n=1}^{\infty} \ (n)^{-4} - (2n)^{-4} = \sum_{n=1}^{\infty} \ (15/16)(n)^{-4}$

Substituting the value of $\sum_{n=1}^{\infty} \ (2n-1)^{-4}$ in the expression given in the question, we get

$\frac{\sum_{n=1}^{\infty} \ (15/16)(n)^{-4}}{ \sum_{n=1}^{\infty} (1/16)(n)^{-4}}$

$=15$

to the above expression add 1 and subtract 1.. hence the expression becomes sigma(n^-4)\sigma((2n)^-4)...fot n=1 to infinite..taking 2^-4 from the denominator ....numerator becomes equal to denominator.. as n goes to infinite... thus we get 1\2^-4 -1 =16-1=15