What are the first few values of this function?

Question

There exists a sequence $a_n$ such that $a_n$ is strictly positive, decreasing, well defined for all $n \in \mathbb{Z}^+$, and obeys the following relationship:

$$\frac{a_n +a_{n+1}}{2} = \frac{1}{n+1}$$

Assume $a_n$ is a rational function of n. What is $a_n$ written in terms of $n$?

I came across this function in studying the Harmonic numbers, $H_x = \sum_{n=1}^x \frac{1}{n}$ The Harmonic numbers can be thought of as the left Riemann sum of the function $1/x$. A visual representation of this sum would be a series of rectangles each with unit width and height $1/x$, being the $x$th rectangle. Imagine the edges of all the rectangles are shifted such that each rectangle now becomes a trapezoid with the same area as the original rectangle, but the edges of the trapezoids form a continuous polygonal line. The heights of these trapezoids would then be the values of $a_n$, leading to the above relationship which follows from the formula for the area of a trapezoid. If $a_n$ is a rational function of $n$, then $\int_0^x a_n dn$ ~ $H_x$

Answer 1

There is no rational function $f$ such that $$ \tag{$\star$} \frac{f(n)+f(n+1)}{2}=\frac{1}{n+1} $$ for all $n\in\mathbb{Z}^+$. An identity between rational functions which holds at the positive integers must hold identically. The right hand side of $(\star)$ has a pole at $n=-1$, so $f(x)$ must have a pole at either $x=-1$ or at $x=0$.

Let $k$ be the largest integer such that $f(x)$ has a pole at $x=k$. This pole cannot cancel a pole of $f(x+1)$, so the left hand side of $(\star)$ has a pole at $x=k$, and hence $k=-1$.

Let $m$ be the smallest integer such that $f(x)$ has a pole at $x=m$. Then $f(x+1)$ has a pole at $x=m-1$ which cannot cancel a pole of $f(x)$, so the left hand side of $(\star)$ has a pole at $x=m-1$, and hence $m-1=-1$, and $m=0$.

We have concluded that the largest integer pole of $f(x)$ is at $x=-1$ and the smallest integer pole is at $x=0$. This is a contradiction.

Answer 2

The recurrence can be rewritten as,

$$(1) \quad a_{n+1}={2 \over {n+1}}-a_n$$

$$\Rightarrow (2) \quad a_n=(-1)^n \cdot a_0+\sum_{k=0}^{n-1} {2 \over {n-k}} \cdot (-1)^k$$

No derivation is given, but $(2)$ is easy enough to check. Comparing to known functions yields,

$$(3) \quad a_n=2 \cdot \Phi(-1,1,n+1)+(a_0-\ln(4)) \cdot (-1)^n$$

Where $\Phi(a,b,c)$ is the Lerch Transcendent. The limit as $n$ approaches infinity doesn't exist. However, it should be noted that the Lerch term goes to $0$.