Prove $\sum _{n=2}^{\infty \:}\frac{1}{n^2+3n-2}=\frac{1}{4}+\frac{\pi \tan \left(\frac{\sqrt{17}\pi }{2}\right)}{\sqrt{17}}$

Question

My sincerest apologies for not being able to provide any context for this problem. I’ve spent about 30 minutes searching for any relevant information or background, but unfortunately, I’ve come up empty-handed.

The sum in question is:

$\sum _{n=2}^{\infty \:}\frac{1}{n^2+3n-2}$

What I tried so far (Basically I failed to do anything):

First, since the roots of the quadratic $n^2+3n-2$ are,

$\frac{-3+\sqrt{17}}{2}$ and $\frac{-3-\sqrt{17}}{2}$

I then attempted partial fraction decomposition on the expression:

$\frac{1}{n^2+3n-2} = \frac{\frac{1}{\sqrt{17}}}{n+\frac{3-\sqrt{17}}{2}} + \frac{\frac{1}{\sqrt{17}}}{n+\frac{3+\sqrt{17}}{2}}$

Thus, the sum can be expressed as:

$\frac{1}{\sqrt{17}}\sum _{n=0}^{\infty }\:\left(\frac{1}{n+\frac{3-\sqrt{17}}{2}}-\frac{1}{n+\frac{3+\sqrt{17}}{2}}\right)$

However, this did not give me the expected result. In fact, I think I may have made things worse by turning a convergent sum into two divergent sums, which obviously isn’t helpful.

I checked Wolfram Alpha for the answer, which gave me the following result:

$\frac{1}{4}+\frac{\pi \tan \left(\frac{\sqrt{17}\pi }{2}\right)}{\sqrt{17}}$

At this point, I thought about adding a term $\frac{1}{2^{n+1}}$ to the original sum in the hope that it would help eliminate the $\frac{1}{4}$ term. However, I wasn’t brilliant enough to come up with anything useful.

So, after spending considerable time on this, I must admit that I made no progress. That is why I am asking the brilliant minds of Stack Exchange for any guidance on how to proceed would be greatly appreciated. Thank you for your time.

Answer 1

Consider first the partial sum

$$S_p=\frac{1}{\sqrt{17}}\sum _{n=0}^{p }\:\left(\frac{1}{n+\frac{3-\sqrt{17}}{2}}-\frac{1}{n+\frac{3+\sqrt{17}}{2}}\right)$$ anr remember $$\sum _{n=0}^{p }\frac{1}{n+a}=\psi (p+a+1)-\psi (a)$$ and the asymptotic $$\psi (p+b)=\log (p)+\frac{2 b-1}{2 p}+\frac{-6b^2+6 b-1}{12 p^2} +O\left(\frac{1}{p^3}\right)$$

Apply is twice and use the series to get $$\left(\psi \left(\frac{3}{2}+\frac{\sqrt{17}}{2}\right)-\psi\left(\frac{3}{2}-\frac{\sqrt{17}}{2}\right)\right)-\frac{\sqrt{17}}{p}+\frac{2 \sqrt{17}}{p^2}+O\left(\frac{1}{p^3}\right)$$ Now, the last simplification $$(\psi \left(\frac{3}{2}+\frac{\sqrt{17}}{2}\right)-\psi\left(\frac{3}{2}-\frac{\sqrt{17}}{2}\right)=\frac{\sqrt{17}}{4}+\pi \tan \left(\frac{\sqrt{17} \pi}{2}\right)$$ which is true looks magic (at least to me). It probably uses some reflection formula.