Series transformation: proving an identity between two sums How does one show that $$\frac{1}{\alpha} + \frac1{2}\frac{1!}{\alpha(\alpha+1)}+ \frac1{3}\frac{2!}{\alpha(\alpha+1)(\alpha+2)}+\ldots = \frac{1}{\alpha^2} + \frac1{(\alpha+1)^2}+\ldots$$
where $\alpha$ is a real number?
I noticed only that if we let $a_k=\frac1{\alpha+k}$, then $\Delta^na_0=\frac{n!}{\alpha(\alpha+1)\ldots(\alpha+n)}$. Is this maybe related to Euler transformation? What can you suggest?
 A: Let $\alpha=p$, then the LHS is nothing but
$$S=\sum_{k=0}^{\infty}(-1)^k \frac{1}{(k+1) {-p \choose k}}$$
Use ${-p \choose k}=(-1)^k{p+k-1\choose k}$, then
$$S=\sum_{k=0}^{\infty} \frac{1}{(k+1) {p+k-1 \choose k}}$$
Next, use $${n \choose k}^{-1}=(n+1)\int_{0}^{1} x^k (1-x)^{n-k}dx$$
$$S=\int_{0}^{1}\sum_{k=0}^{1} \frac{p+k}{k+1} x^k (1-x)^{p-1} dx$$
$$S(p)=\int_{0}^{1}\left(\sum_{k=0}^{\infty} \frac{p-1}{k+1} x^k (1-x)^{p-1}+ \sum_{k=0}^{\infty} x^k (1-x)^{p-1}\right) dx.$$
$$S(p)=-(p-1)\int_{0}^{1} \frac{\ln(1-x) (1-x)^{p-1}}{x} dx+\int_{0}^{1} (1-x)^{p-2} dx$$
Using the definition of Polygamma function $\psi^m(z)$
$$S(p)=(p-1) \psi^{1}(p)+\frac{1}{p-1}$$
Which in terms of Zeta function can be written as
$$S(p)=(p-1) \zeta(2,p)+\frac{1}{p-1}.$$
Finally, we can write
$$S(p)=(p-1) \sum_{k=0}^{\infty} \frac{1}{(p+k)^2}+\frac{1}{p-1}, ~p>1.$$
For instance we get $S(1.1)=10.1433.., S(3/2)=\pi^2/4, S(2)=\pi^2/6, S(5/2)= 3(\pi^2-8)/4, S(3)=(\pi^2-6)/3.$$
These results have also been checked at Mathematica numerically.
For Polygamma functions see:
https://en.wikipedia.org/wiki/Polygamma_function#:~:text=In%20mathematics%2C%20the%20polygamma%20function,on%20%E2%84%82%20%5C%20%E2%88%92%E2%84%950.
