Let $f(n)=\frac{3^{2n}}{3^{2n}+3},$ calculate $\sum_{k=1}^{2018} f \left(\frac{k}{2019}\right)$

Question

Let $f(n)=\frac{3^{2n}}{3^{2n}+3}$. Find $$f \left(\frac{1}{2019}\right)+f\left(\frac{2}{2019}\right)+f\left(\frac{3}{2019}\right)+...+f\left(\frac{2018}{2019}\right).$$

Answer 1

Note that $$f(x)=\frac{3^{2n}}{3^{2n}+1}=1-\frac1{3^{2n}+1}$$ so $$S=\sum_{k=1}^{2018}f\left(\frac k{2019}\right)=2018-\sum_{k=1}^{2018}\frac1{3^{2k/2019}+1}$$

We can obtain upper and lower bounds.

$$3^{2k/2019}+1<10\Leftarrow\frac{2k}{2019}<2\Leftarrow k<2019$$ so $$S<2018-2018\cdot\frac1{10}\implies S<1816.2$$ Also, $$3^{2k/2019}+1>2\Leftarrow k>0$$ so $$S>2018-2018\cdot\frac12\implies S>1009$$

Hence a rather wide interval for $S$ is $$1009<S<1816.2$$

Answer 2

Hint

$$f(n)=\frac {3^{2n}}{3^{2n}+3}$$

We see that $$f\left(\frac {1}{2019}\right)+f\left(\frac {2018}{2019}\right)=1$$ And similarly $$f\left(\frac {2}{2019}\right)+f\left(\frac {2017}{2019}\right)=1$$

And so on until $$f\left(\frac {1009}{2019}\right) +f\left(\frac {1010}{2019}\right)=1 $$

Hence the sum would simply be $$1+1+1+....(\text { 1009 times})=1009$$

Answer 3

More generally, if $f(x)=\frac{a^{2x}}{a^{2x}+a}$ with $a>0$ then $$f(1-x)=\frac{a^{2(1-x)}}{a^{2(1-x)}+a}=\frac{a\cdot a^{-(2x-1)}}{a\cdot a^{-(2x-1)}+a}=\frac{a}{a+a^{2x}}=1-\frac{a^{2x}}{a+a^{2x}}=1-f(x).$$ Hence for any positive integer $n$, $$\sum_{k=1}^{n-1}f\left(\frac{k}{n}\right)=\frac{1}{2}\left(\sum_{k=1}^{n-1}f\left(\frac{k}{n}\right)+\sum_{k=1}^{n-1}f\left(1-\frac{k}{n}\right)\right)=\frac{\sum_{k=1}^{n-1}\left(f\left(\frac{k}{n}\right)+f\left(1-\frac{k}{n}\right)\right)}{2}=\frac{n-1}{2}.$$ Notice that the sum does not depend on $a$.