# 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.)

• What is $3.5$ ? I mean, the number $7/2$ or the multiplication $15$ ? – user1971 Oct 11 '15 at 15:09
• It's not an integer, if $3.5 = 7/2$. – Patrick Stevens Oct 11 '15 at 15:09
• With a computer I get the value 2005.736 if 3.5=7/2 and 1984.4776 if 3.5=15. Just check the question again. – Kartik Oct 11 '15 at 15:19
• @Kartik I make it 559.51119983 in the first case. Mathematica: g[t_] := 3 t/(3 t + 7/2); N[Sum[g[i/2015], {i, 1, 2014}], 50] – Patrick Stevens Oct 11 '15 at 15:21
• @PatrickStevens Yes, sorry thats right. Bug in program. 559.511 in the first case and 178.026 in the second. – Kartik Oct 11 '15 at 15:25

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.

• And this put me over 30k points. I am proud of this answer. – marty cohen Oct 11 '15 at 23:47
• A whiff of Gauss... nicely done. – J. M. is a poor mathematician Oct 12 '15 at 4:47
• Thank you. Upvoted, of course. – marty cohen Oct 12 '15 at 4:48

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.

• By the way, Mathematica says that the difference between the two expressions is approx 0.2308. The sum has value approx 559.511199837. – Patrick Stevens Oct 11 '15 at 15:19
• @PatrickStevens: $559.742$ or $559.511$, we still don't have an integer. – Jack D'Aurizio Oct 11 '15 at 15:21
• Agreed. Just wanted to quantify "the value is about". – Patrick Stevens Oct 11 '15 at 15:22
• nice answer! Right now I wonder, is there any fast way to see that $6\cdot 2001+5\cdot 7\cdot 13\cdot 31$ is prime? Is this actually the only prime in the denominator? – user190080 Oct 11 '15 at 15:22
• @user190080: actually there are plenty of primes among the denominators. I just checked a large denominator for primality, since that gives in a very straightforward way that the sum cannot be an integer, like in the classical proof of $H_n\not\in\mathbb{Z}$ for any $n>1$. – Jack D'Aurizio Oct 11 '15 at 15:24

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.

\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}

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

$$\sum_{k = 1}^{2014} g\left(\frac{k}{2015}\right) = \sum_{k = 1}^{1007} \left ( g\left(\frac{k}{2015} \right ) + g\left(\frac{2015 - k}{2015} \right ) \right ) = 1007.$$