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.)

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.
