Show that this sum is an integer. 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.)
 A: 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.
A: 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.
A: $$\begin{align}
&g(t)=\frac {3^t}{3^t+3^{\frac 12}}=\frac {3^{t-\frac12}}{3^{t-\frac 12}+1}\\
\Rightarrow \qquad &g\left(\frac12+u\right)+g\left(\frac12-u\right)=\frac {3^u}{3^u+1}+\frac {3^{-u}}{3^{-u}+1}=\frac {3^u}{3^u+1}+\frac {1}{1+3^u}=1\\\\
\sum_{r=1}^{2014}g\left(\frac r{2015}\right)
&=\sum_{r=1}^{1007}g\left(\frac r{2015}\right)+\sum_{r=1008}^{2014}g\left(\frac r{2015}\right)\\
&=\sum_{i=1}^{1007}g\left(\frac{1007\frac12-(i-\frac12)}{2015}\right)+g\left(\frac{1007\frac12+(i-\frac12)}{2015}\right)\\
&=\sum_{i=1}^{1007}g\left(\frac12-\frac{i-\frac12}{2015}\right)+g\left(\frac12+\frac{i-\frac12}{2015}\right)\\
&=\sum_{i=1}^{1007}1\\
&=1007\qquad\blacksquare\end{align}$$
A: 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 
$$\sum_{k = 1}^{2014} g\left(\frac{k}{2015}\right) = \sum_{k = 1}^{1007} \left ( g\left(\frac{k}{2015} \right ) + g\left(\frac{2015 - k}{2015} \right ) \right ) = 1007.$$
A: 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.
