Prove that $\frac{1}{a^2}+\frac{1}{(a+1)^2}+\frac{1}{(a+2)^2}+\dotsm\infty=\frac{1}{a}+\frac{1}{2a(a+1)}+\frac{2!}{3a(a+1)(a+2)}+\dotsm\infty$ Question:-  Prove that $$\frac{1}{a^2}+\frac{1}{(a+1)^2}+\frac{1}{(a+2)^2}+\dotsm\infty=\frac{1}{a}+\frac{1}{2a(a+1)}+\frac{2!}{3a(a+1)(a+2)}+\dotsm\infty$$
Nothing is mentioned in question about nature of $a$
I write it in summation form,but I got stuck and unable to proceed further.
$$\sum_{k=0}^{\infty}\frac{1}{(a+k)^2}=\sum_{n=0}^{\infty}\frac{n!}{(n+1)\prod_{k=0}^{n}(a+k)}$$
Then I take all the terms to LHS in hope that terms may cancel out each other to give zero but that also doesn't help me since with each term degree of both numerator and denominator increases.
Can anybody help me to Prove the result!!
 A: Suppose that $\Re(a)>0$. Then we have
\begin{align*}
\sum\limits_{k = 0}^\infty  {\frac{1}{{(a + k)^2 }}} & = \sum\limits_{k = 0}^\infty  {\int_0^{ + \infty } {e^{ - (a + k)t} tdt} }  = \int_0^{ + \infty } {e^{ - at} t\sum\limits_{k = 0}^\infty  {e^{ - kt} } dt}  = \int_0^{ + \infty } {e^{ - at} \frac{t}{{1 - e^{ - t} }}dt} 
\\ & \mathop  = \limits^{t =  - \log s} \int_0^1 {s^{a - 1} \frac{{ - \log s}}{{1 - s}}ds}  = \int_0^1 {s^{a - 1} \frac{{ - \log (1 + (s - 1))}}{{1 - s}}ds} 
\\ &  = \int_0^1 {s^{a - 1} \sum\limits_{k = 1}^\infty  {( - 1)^{k - 1} \frac{{(s - 1)^{k - 1} }}{k}} ds}  = \sum\limits_{k = 1}^\infty  {\frac{1}{k}\int_0^1 {s^{a - 1} (1 - s)^{k - 1} ds} } 
\\ & = \sum\limits_{k = 1}^\infty  {\frac{1}{k}\frac{{\Gamma (k)\Gamma (a)}}{{\Gamma (a + k)}}}  = \sum\limits_{k = 1}^\infty  {\frac{{(k - 1)!}}{{ka(a + 1) \cdots (a + k - 1)}}} .
\end{align*}
Remark: The original sum is actually convergent provided that $a\neq 0,-1,-2,\ldots$ and its sum is the trigamma function
$$
\psi _1 (a) = \frac{{d^2 }}{{da^2 }}\log \Gamma (a).
$$
The series form after the transformations is convergent only when $\Re(a)>0$. It is called a factorial series expansion.
