Does the series $\sum\limits_{k=1}^{\infty}a_k$ converge if $a_k=\frac{k^2-k+c}{k^2}a_{k-1}$ for $k=2,3,..$ and $c$ is some constant?

Question

If $c=0$ then the series obviously diverges (as it is then identical to harmonic series). I think it then follows by direct comparison test that the series also diverges for $c>0$.

I therefore guess that the series is divergent for any $c$ but can't figure out how to prove it precisely.

Answer 1

The series does not necessarily diverge for all $c$. Consider any $c=n-n^2$ for some natural $n$ and you will find that $a_k=0$ for all $k\ge n$.

For $c$ not of the above form, however, the series does in fact, diverge. Note that we have

\begin{align}a_k&=a_2\prod_{n=2}^k\frac{n^2-n+c}{n^2}\\&=\frac{a_2}k\prod_{n=2}^k\frac{n^2-n+c}{n^2-n}\\&=\frac{a_1}k\exp\sum_{n=2}^k\ln\left(1+\frac c{n^2-n}\right)\\&\sim\frac{a_1}k\exp\sum_{n=2}^k\frac c{n^2-n}\\&=\frac{a_1}k\exp\left(c-\frac ck\right)\end{align}

which diverges by the limit comparison test.

Answer 2

Similar to Simply Beautiful Art's answer, I think that the sum cannot converge.

Using $$a_k=\frac{k^2-k+c}{k^2}a_{k-1}\qquad \text{with} \qquad a_1=A$$ using Pochhammer symbols $$a_k=A\frac{ \left(\frac{1}{2} \left(\sqrt{1-4 c}+1\right)+1\right)_{k-1} \left(\frac{3}{2}-\frac{1}{2} \sqrt{1-4 c}\right)_{k-1}}{\left((2)_{k-1}\right){}^2}$$ Using Taylor expansion for large values of $k$ $$\log(a_k)=\log \left(\frac{A}{\Gamma \left(\frac{1}{2} \left(\sqrt{1-4 c}+1\right)+1\right) \Gamma \left(\frac{3}{2}-\frac{1}{2} \sqrt{1-4 c}\right)}\right)+\log \left(\frac{1}{k}\right)-\frac{c}{k}+\frac{c^2}{6 k^3}+O\left(\frac{1}{k^5}\right)$$ Applying again for $a_{k+1}$ and continuing with Taylor series $$\log(a_{k+1})-\log(a_k)=-\frac{1}{k}+\frac{c+\frac{1}{2}}{k^2}-\frac{c+\frac{1}{3}}{k^3}+O\left(\frac{1}{k^4}\right)$$ Continuing with Taylor $$\frac{a_{k+1} } {a_{k}}=e^{\log(a_{k+1})-\log(a_k)}=1-\frac{1}{k}+\frac{c+1}{k^2}+O\left(\frac{1}{k^3}\right)$$