I have to show that
$$g\left(\frac{1}{2015}\right) + g\left(\frac{2}{2015}\right) +\cdots + g\left(\frac{2014}{2015}\right) $$ is an integer. Here $g(t)=\dfrac{3^t}{3^t+3^{1/2}}$.
I tried to solve it using power series, but I can not finish with any convincing argument to said that the sum is an integer. (Also the use of a computer isn't allowed.)
On
This is the answer to the first version of the question:
We want to compute:
$$ \sum_{k=1}^{2014}\frac{\frac{3k}{2015}}{\frac{3k}{2015}+\frac{7}{2}}=\sum_{k=1}^{2014}\frac{6k}{6k+5\cdot 7\cdot 13\cdot 31} $$ but that number cannot be an integer because $6\cdot 2001+5\cdot 7\cdot 13\cdot 31$ is a prime.
Anyway, the value of the LHS is about $2015\int_{0}^{1}\frac{3x}{3x+\frac{7}{2}}\,dx = 2015\left(1-\frac{7}{6}\log\frac{13}{7}\right)$.
We may also estimate the difference between our sum and $2015\left(1-\frac{7}{6}\log\frac{13}{7}\right)$ through the Hermite-Hadamard inequality, since $\frac{3x}{3x+\frac{7}{2}}$ is a concave function on $[0,1]$. That gives another way for proving that our sum is not an integer, since it gives that our sum is between $559.4$ and $559.8$.
This is the answer to the second version of the question. If $$ g(t) = \frac{3^t}{3^t+3^{1/2}} $$ we have: $$ g(t)+g(1-t) = \frac{3^t}{3^t+3^{1/2}}+\frac{3^{1-t}}{3^{1-t}+3^{1/2}}=\frac{1}{1+3^{1/2-t}}+\frac{1}{1+3^{t-1/2}}=\color{red}{1}$$ hence the claim is trivial.
On
By a rigid translation you can see that $g$ is an odd function around the point $(\frac 1 2, \frac 1 2)$:
$$ \begin{align} f(x) &= g(x+\frac 1 2) - \frac 1 2 \\ &= \frac{3^{x + 1/2}}{3^{x + 1/2} + 3^{1/2}} - \frac 1 2 \\ &= \frac{3^x 3^{1/2}}{3^x 3^{1/2} + 3^{1/2}} - \frac 1 2 \\ &= \frac{3^x 3^{1/2}}{(3^x+1) 3^{1/2}} - \frac 1 2 \\ &= \frac{3^x}{3^x+1} - \frac 1 2 \\ &= \frac{2(3^x)-(3^x+1)}{2(3^x+1)} \\ &= \frac{1}{2}\frac{3^x-1}{3^x+1} \\ &= \frac{1}{2}\frac{e^{x\log(3)}-1}{e^{x\log(3)}+1} \\ &= c_1 \tanh(c_2x) \ \ \text{ with $c_1 = \frac 1 2$ and $c_2 = \log(3)$ } \end{align} $$
The last formula is the hyperbolic tangent of a multiple of $x$, so it's symmetric with respect to the origin. From this, it follows that the $f(x) + f(-x) = 0$, which, after the transformation, means $g(x) + g(1-x) = 1$.
From this the claim follows.
On
$$\begin{align} &g(t)=\frac {3^t}{3^t+3^{\frac 12}}=\frac {3^{t-\frac12}}{3^{t-\frac 12}+1}\\ \Rightarrow \qquad &g\left(\frac12+u\right)+g\left(\frac12-u\right)=\frac {3^u}{3^u+1}+\frac {3^{-u}}{3^{-u}+1}=\frac {3^u}{3^u+1}+\frac {1}{1+3^u}=1\\\\ \sum_{r=1}^{2014}g\left(\frac r{2015}\right) &=\sum_{r=1}^{1007}g\left(\frac r{2015}\right)+\sum_{r=1008}^{2014}g\left(\frac r{2015}\right)\\ &=\sum_{i=1}^{1007}g\left(\frac{1007\frac12-(i-\frac12)}{2015}\right)+g\left(\frac{1007\frac12+(i-\frac12)}{2015}\right)\\ &=\sum_{i=1}^{1007}g\left(\frac12-\frac{i-\frac12}{2015}\right)+g\left(\frac12+\frac{i-\frac12}{2015}\right)\\ &=\sum_{i=1}^{1007}1\\ &=1007\qquad\blacksquare\end{align}$$
On
Notice that $g (x) = g (1 - x)$. Indeed
$$g(x)+g(1-x) = \frac {1} {1 + 3^{1/2-x}} + \frac {1} {1+3^{x-1/2}} = 1$$.
Then we conclude by
$$\sum_{k = 1}^{2014} g\left(\frac{k}{2015}\right) = \sum_{k = 1}^{1007} \left ( g\left(\frac{k}{2015} \right ) + g\left(\frac{2015 - k}{2015} \right ) \right ) = 1007.$$
Want $\sum_{k=1}^{n-1} g\left(\frac{k}{n}\right) $ where $n$ is odd and $g(t)=\dfrac{3^t}{3^t+3^{1/2}} $.
$g(k/n) =\dfrac{3^{k/n}}{3^{k/n}+3^{1/2}} $.
$\begin{array}\\ g(k/n)+g((n-k)/n) &=\dfrac{3^{k/n}}{3^{k/n}+3^{1/2}}+\dfrac{3^{(n-k)/n}}{3^{(n-k)/n}+3^{1/2}}\\ &=\dfrac{3^{k/n}(3^{(n-k)/n}+3^{1/2})+3^{(n-k)/n}(3^{k/n}+3^{1/2})}{(3^{k/n}+3^{1/2})(3^{(n-k)/n}+3^{1/2})}\\ &=\dfrac{(3+3^{(k/n)+(1/2)})+(3+3^{(n-k)/n+(1/2)})}{3+3^{1/2}(3^{k/n}+3^{(n-k)/n})+3}\\ &=\dfrac{6+3^{1/2}(3^{k/n}+3^{(n-k)/n})}{6+3^{1/2}(3^{k/n}+3^{(n-k)/n})}\\ &= 1\\ \end{array} $
Wow! This was completely unexpected.
Since the sum of paired terms is one, if $n$ is odd, and there are $\frac{n-1}{2}$ pairs the sum is $\frac{n-1}{2}$.
It looks like any number can be substituted for $3$ and this will work.