$\frac{1+\frac{1}{2^{11}}+\frac{1}{3^{11}}+\frac{1}{4^{11}}+...}{1-\frac{1}{2^{11}}+\frac{1}{3^{11}}-\frac{1}{4^{11}}+...}$

Question

I'm not sure how to solve this summation and would like assistance. I believe the answer is ||$\frac{1024}{1023}$|| according to Wolfram. I don't know what the proper approach is (creating two summations and somehow combining or common denominators..?).

Answer 1

For $s>1$, we have $$ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}= \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s} + \cdots $$This is often known (by a slight abuse of terminology) as the zeta function. Similarly, for $s>0$ we have $$ \eta (s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}= \frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s} + \cdots $$Note that if we subtract them we have $$ \zeta(s)-\eta(s) = \frac{2}{2^{s}}+\frac{2}{4^s}+\frac{2}{6^s}+\frac{2}{8^s}+\cdots $$ $$ = \frac{2}{2^s}\left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s} + \cdots\right) = 2^{1-s}\zeta(s), $$or $\eta(s) = (1-2^{1-s})\zeta(s)$. This brings us back to your question. The numerator is clearly $\zeta(11)$ and the denominator is $2-\eta(11)$. As written, this doesn't simplify nicely. If, however, the pattern in the denominator were $1-2^{-11}+3^{-11}-4^{-11}+\cdots$, then we'd have $\zeta(11)/((1-2^{-10})\zeta(11))=1024/1023$, as you've suggested. Hope that helps!

Answer 2

I will assume that the +1 in the denominator is a -1 and go on, we can write $\frac{1}{1}+\frac{1}{2^{11}}+\frac{1}{3^{11}}+.......=S$

now we can take the even terms to one side and express it as

$\frac{1}{1}$+$\frac{1}{3^{11}}$+$\frac{1}{5^{11}}$=S +$\frac{1}{2^{11}}(S)$

$\frac{1}{1}+\frac{1}{3^{11}}+\frac{1}{5^{11}}=S(\frac{2^{11}-1}{2^{11}})$

now the equation in the question becomes

$\frac{S}{S-2(S(\frac{2^{11}-1}{2^{11}})}$ which gives the answer $\frac {1024}{1023}$