Show that this sum is an integer.

Solution 1:

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.

Solution 2:

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.