Is it possible to determine what is the arithmetic mean of the harmonic series where $n$ and $-n$ are added and divided by two in this manner:
$$f(n) = \frac{ (\frac{3}{4})^n \times 2^{⌈-n \times log_2(\frac{3}{4})⌉} + (\frac{3}{4})^{-n} \times 2^{⌈n \times log_2(\frac{3}{4})⌉} }{2}$$
For each positive plus negative whole number $n$ we have an arithmetic mean. See values in the continuous graph below:
Now I want to sum them and take an average something like:
$$\frac{\sum_{n=1}^{x}{f(n)}}{x}$$
Does this series converge, if x is infinity?
My preliminary spreadsheet calculation gives a value somewhere near to $(6/5)^2 = 1.44$ if $n$ is $20$. Or, maybe a cuberoot of three? See row 20 in the picture. But I'm missing tools to determine if this equation has a limit when $n$ is a whole number approaching infinity?
Possible limit equation
$$ \lim_{h \to \infty}{\frac{1}{h}\sum_{n=1}^{h}{ \frac{ (\frac{3}{4})^n \times 2^{⌈-n \times log_2(\frac{3}{4})⌉} + (\frac{3}{4})^{-n} \times 2^{⌈n \times log_2(\frac{3}{4})⌉} }{2} }} $$
With Wolfram Alpha I got the summation formula simplified to:
$$g(n) = \frac{2^{-\frac{n\log\frac{4}{3}}{\log2}\pmod1} + 2^{\frac{n \log\frac{4}{3}}{\log2}\pmod1}}{2}$$
The distribution of the first 500 values in graph looks like this:
Spreadsheet
Col 7 is (col3+col4)/2 corresponding equation in my question. Cols 3, 4, 7 are points (x, y) but x only is applicable now. Col 1 and 2 are positive and negative $n$.
Added references
This seems to relate to Cesaro mean:
I believe the answer is $\frac12\int_0^1 (2^x+2^{1-x})dx = \frac{1}{\log2}\approx 1.44269504$. This follows from the equidistribution theorem (look for Khinchin's integral and apply to the functions $f_1(x)=2^x$ and $f_2(x)=2^{1-x}$).
With some more steps, we have $(\frac34)^n=2^{n \log_2 \frac34}$ so $$ (\frac34)^n 2^{\lceil -n\log_2 \frac34 \rceil} = 2^{n\log_2 \frac34+\lceil -n\log_2 \frac34 \rceil} $$ and $$(\frac34)^{-n} 2^{\lceil n\log_2 \frac34 \rceil} = 2^{-n\log_2 \frac34+\lceil n\log_2 \frac34 \rceil}. $$ Now for $x\notin\mathbb{Z}$, $x+\lceil -x\rceil = x \bmod 1$ and $-x+\lceil{x}\rceil=(1-x) \bmod 1=1+(-x \bmod 1)$.
Set $\alpha=\log_2{\frac34} = \log_23-2$. Then $\alpha$ is irrational because otherwise, you'd have $3=2^{\frac pq}$ for integers $p$ and $q$, so $3^q=2^p$, clearly impossible.
So we are calculating $$\frac1{2N} \sum_{n=1}^N \left(2^{(n\alpha \bmod 1)}+ 2^{1+(-n\alpha \bmod 1)}\right) $$
Khinchin's identity now says that by the irrationality of $\alpha$, the limit is $$\frac12\int_0^1 \left(2^{x}+2^{1-x}\right) dx = \frac1{2\ln 2}\left[2^x-2^{1-x} \right]_0^1 =\frac1{2\ln2}(2-1-1+2)=\frac{1}{\ln 2}. $$