Question

Let $A=16-4+2^{-4}+3^{-4}+4^{-4}+...$ and $B=1^{-4}+3^{-4}+5^{-4}+7^{-4}+...$. The ratio $\frac{A}{B}$ in the lowest form is

• A. $\frac{16}{15}$
• B. $\frac{15}{14}$
• C. $\frac{15}{16}$
• D. $\frac{13}{12}$

Answer 1

The correct option is A $\frac{16}{15}$

$A=16-4+2^{-4}+3^{-4}+4^{-4}+....$

$A=(11+1^{-4}+2^{-4}+....)$

$\sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{90}$

$A=11+\frac{{\pi }^4}{90}$

$B=1^{-4}+3^{-4}+5^{-4}+....$

$=(1^{-4}+2^{-4}+3^{-4}+....\infty )-(2^{-4}+4^{-4}+6^{-4}+....)$

$B=\frac{{\pi }^4}{90}-\sum _{n=1}^{\infty }\frac{1}{{\left(2n\right)}^4}$

$=\frac{{\pi }^4}{90}-\frac{1}{16}\sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{90}-\frac{{\pi }^4}{90\times 16}$

$=\left(\frac{15}{16}\right)\frac{{\pi }^4}{90}$

$\frac{A}{B}=\frac{11+\frac{{\pi }^4}{90}}{\left(\frac{15}{16}\right)\frac{{\pi }^4}{90}}=\left(\frac{16}{15}\right)[\frac{90\times 11}{{\pi }^4}+1]$

Lowest form will be $\frac{16}{15}$