<s>[INST] The limit of the ratio of polygamma functions I want to calculate this quantity:
$$\lim_{x \rightarrow \infty}\frac{\Psi_1 (x)}{\Psi_1 (x + y)}$$
 where
$$\Psi_1 (x)=\frac{d^2}{dx^2}\log \Gamma (x)=\sum_{k=0}^{\infty}\frac{1}{(x+k)^2}. $$
I guess it is $1$, but I am not sure about my proof.
My proof is following:
Let $\epsilon > 0$ be given. Since $\Psi_1 (x)$ is convergent on $(0, \infty)$ and decreasing, for any $x, y >0$ there exist $K=K(\epsilon) < \infty$ such that $\sum_{k=K+1}^{\infty} \frac{1}{(x + y + k)^2} < \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \leq \epsilon$. Then, we have
\begin{align*}
       &\frac{\Psi_1 (x)}{\Psi_1 (x + y)} \leq \Bigg( \sum_{k=0}^{K} \frac{1}{(x + k)^2} + \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
       %&\leq \Bigg( \sum_{k=0}^{K} \frac{1}{(x + k)^2} - \frac{1}{(x + y + k)^2}  \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg) + 2\epsilon\\
       &\leq \Bigg( \frac{K+1}{x^2} + \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
       &= \Bigg( \frac{K+1}{x^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)
       + \Bigg( \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
       & \leq \Bigg( \frac{K+1}{x^2} \Bigg) \Bigg/ \Bigg( \frac{K+1}{(x + y + K)^2} \Bigg)
       + \Bigg( \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
       & = \frac{(x + y + K)^2}{x^2}
       + \Bigg( \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg).
\end{align*}
In the last line, the first term goes to $1$ as $x \longrightarrow \infty$ for any fixed $K$. For the second term, since we can choose arbitrarily large $K$ and $\Psi_1$ is convergent, it goes to $0$ as $K \longrightarrow \infty$ for any fixed $x>0$. Thus, we have $\frac{\Psi_1 (x)}{\Psi_1 (x + y)} \leq 1$ as $x \longrightarrow \infty$.
On the other hand, since $\Psi_1(x)$ is decreasing in $x$,
\begin{equation*}
    \frac{\Psi_1 (x)}{\Psi_1 (x + y)} \geq 1 \text { for any } x, y >0.
\end{equation*}
Thus,
\begin{equation*}
    \frac{\Psi_1 (x)}{\Psi_1 (x + y)} \longrightarrow 1  \text { as } x \longrightarrow \infty.
\end{equation*}
Am I correct? I think I am cheating somewhere. I can't convince myself.
 [/INST] Why not to use the asmptotics
$$\Psi_1 (z)=\frac{1}{z}+\frac{1}{2 z^2}+\frac{1}{6
   z^3}+O\left(\frac{1}{z^5}\right)$$ Apply it twice and continue with Taylor series to get
$$\frac{\Psi_1 (x)}{\Psi_1 (x + y)}=1+\frac{y}{x}+\frac{y}{2 x^2}+\frac{y}{12
   x^3}+O\left(\frac{1}{x^4}\right)$$
Trying with $x=100$ and $y=10$, the "exact" value is $1.100500898$ while the above truncated series gives $\frac{1320601}{1200000}=1.100500833$.
</s>