# Sum of the fractional series

I've got a series and I wanted to calculate the sum. Thanks for help. The series is

$${\sum _{n=1}^{\infty } \frac{a_n}{b_n}}$$

Lets suppose that

$${\sum _{n=1}^{\infty }{a_n}}=C$$

$${\sum _{n=1}^{\infty }{b_n}}=D$$

Can I calculate with known constants $C$ and $D$ this series?

• Have you tried anything? – Matthew Conroy Jan 19 '18 at 1:31
• This is not any where near enough information. There are infinitely many series whose sum is $C$ for any given $C$. For simple example, let $a_n = (1,1,0,0,0,\dots,)$ and let $b_n=(\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots,\frac{1}{2^k},\dots)$. We would have $\sum a_n = 2$ and $\sum b_n = 1$ while $\sum \frac{a_n}{b_n}=6$. Compare this to if we replace the first sequence as $a'_n = (2,0,0,\dots)$, which has the same sum as before: $\sum a'_n=2$, but this time $\sum\frac{a'_n}{b_n}=4$, a different result than before. – JMoravitz Jan 19 '18 at 1:52
• If there was such a result that you are looking for, getting different answers should have been impossible. Since we got different answers for the same initial conditions that implies there is no way to calculate it knowing only $C$ and $D$. We would need to know more about the specific sequences $a_n$ and $b_n$. – JMoravitz Jan 19 '18 at 1:54

This is impossible in general form. Consider $a_n=\frac{1}{4^n}$ and $b_n=\frac{1}{2^n}-\frac{1}{3^n}$. Therefore $C=\frac{1}{3}$ and $D=\frac{1}{2}$ while there is no known closed form for $\sum_{n=1}^{\infty}\frac{a_n}{b_n}=\sum_{n=1}^{\infty}\frac{6^n}{12^n-8^n}$