I know that $\sum_{n\geq 1} \frac{1}{n^2}=\pi^2/6$, is there a simple way to get an explicit closed form $\sum \frac{1}{an^2+bn+c}$, where $a,b,c$ are integers, $a\neq 0$?
I would like to evaluate the sum  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$  Here is my attempt:    Letting    $$f(z)=\frac{1}{az^2+bz+c}$$  The poles of $f(z)$ are located at   $$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$  and  $$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$  Then  $$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$  Using L'Hopital's rule. Continuing, we have the limit is  $$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$  For $z_0 \ne 0$  Similarly, we find  $$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$  Then   $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=  -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$  Then we have  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$    Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum?
My script (should) acts differently, depending on the presence of the data in the input stream. So I can invoke it like this:  $ my-script.sh   or:  $ my-script.sh &lt;&lt;-MARK     Data comes...     ...data goes. MARK   or:  $ some-command | my-script.sh   where two last cases should read the data, while first case should notice the data is missing and act accordingly.  The crucial part (excerpt) of the script is:  #!/bin/bash local myData; read -d '' -t 0 myData;  [ -z "${myData}" ] &amp;&amp; {     # Notice the lack of the data. } || {     # Process the data. }   I use read to read input data, then option -d '' to read multiple lines, as this is expected, and the -t 0 to set timeout to zero. Why the timeout? According to  (typing left unchanged; bold is mine):     -t timeout time out and return failure if a complete line of input is               not read withint TIMEOUT seconds.  The value of the TMOUT               variable is the default timeout.  TIMEOUT may be a               fractional number.  If TIMEOUT is 0, read returns success only               if input is available on the specified file descriptor.  The               exit status is greater than 128 if the timeout is exceeded   So I in case 2 and 3 it should read the data immediately, as I understand it. Unfortunately it doesn't. As -t can take fractional values (according to above man page), changing read line to:  read -d '' -t 0.01 myData;   actually reads the data when data is present and skips it (after 10ms timeout) if it is not. But it should also work when TIMEOUT is set to real 0.  Why it actually doesn't? How can this be fixed? And is there, perhaps, alternative solution to the problem of "act differently depending on the presence of the data"?  UPDATE  Thanks to @Isaac I found a misleading discrepancy between quoted on-line version and my local one (normally I do not have locale set to en_US, so help read gave me translation which I couldn't paste here, and looking up for on-line translation was faster than setting new env---but that caused the whole problem).  So for 4.4.12 version of Bash it says:     If TIMEOUT is 0, read returns               immediately, without trying to read any data, returning               success only if input is available on the specified               file descriptor.   This gives a little bit different impression than "If TIMEOUT is 0, read returns success only if input is available on the specified file descriptor"---for me it implied actually an attempt to read the data.  So finally I tested this and it worked perfectly:  read -t 0 &amp;&amp; read -d '' myData;   The meaning: see if there's anything to read and if it succeed, just read it.  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How to use mysql JOIN without ON condition? proof that $f'$ is continuous at $a$ Let $f:I \rightarrow \mathbb{R}$ derivable in set I. if $\lim x_n = \lim y_n =a \in I$ with $x_n \neq y_n$ and $\lim \frac{f(y_n)-f(x_n)}{y_n - x_n} = f'(a)$ proof that $f'$ is continuous at $a$  my attempt $x_n \neq y_n$ so by MVT $\exists z_n \in I$ like $x_n \geq z_n \geq y_n$ with $\frac{f(y_n)-f(x_n)}{y_n - x_n}=f'(z_n)$ for each $x_n$ and  $y_n$ then $f'(a) = \lim\frac{f(y_n)-f(x_n)}{y_n - x_n}= \lim f'(z_n)$   Using $\lim x_n = \lim y_n = a$ and $x_n \geq z_n \geq y_n \Rightarrow z_n \rightarrow a$  if $ \lim_{x\rightarrow a} f'(x) \neq f'(a)$ then $\exists \epsilon &gt; 0$ such $\forall n\in \mathbb{N}$, $z_n \in [x_n,y_n]$ $0 &lt; |z_n - a| &lt; \frac{1}{n}$ and $|f'(z_n) - f'(a)| \geq \epsilon$. 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