on the sum of an infinite series

Question

Got stuck with this series: $$ \sum_{k=0}^\infty \frac{1}{(\theta+2+k)(\theta+1)^{k+1}} $$ which should be equal to $$ \int_0^1 \frac{t^{\theta+1}}{\theta+1-t}\textrm{d}t $$ But why? Which is the missing passage?

Answer 1

Differentiation and integration will help, you have to somehow construct a geometric series and choosing a function, that upon differentiation, will eliminate the extra term, you will succeed.

Consider a function $$f(t)=\frac{t^{\theta+1+k+1}}{(\theta+1+k+1)(\theta+1)^{k+1}}$$ Differentiate $$f'(t)=\frac{t^{\theta+k+1}}{(\theta+1)^{k+1}}=t^\theta \left(\frac{t}{\theta+1}\right)^{k+1}$$ Sum $$\sum_{k=0}^\infty f'(t)=t^{\theta}\frac{t/(\theta+1)}{1-t/(\theta+1)}=\frac{t^{\theta+1}}{\theta+1-t}$$ Integrate back and take into account that the original sum was $\sum_{k=0}^\infty f(1)$.

This is a very general trick regarding the generating functions. All sums that differ from geometric sum by polynomial factors in $k$, can be solved with this method.