A question about the digamma function and a recursion

The recursion is $T_n=\frac{b}{n+b}+T_{n-1}$ and $T_1=1$. When expanding the recursion we have $T_n=\frac{b}{b+n-1}+\frac{b}{b+n-2}+...+\frac{b}{b+1}+1$. However, the solution shows $T_n=b({\psi _0}(b + n) - {\psi _0}(b))$ where ${\psi _0}$ is the digamma function. The solution indicates ${\psi _0}(b + n) - {\psi _0}(b)=\frac{1}{b+n-1}+\frac{1}{b+n-2}+...+\frac{1}{b+1}+\frac{1}{b}$.

For instance, when $n=4,b=1$, ${\psi _0}(5) - {\psi _0}(1) = \frac{{25}}{{12}}$, and $\frac{1}{b+n-1}+\frac{1}{b+n-2}+...+\frac{1}{b+1}+\frac{1}{b}=\frac{25}{12}$ as well, so I believe the solution is correct.

I am not familiar with digamma function so I checked Wikipedia, but I still cannot understand this equation. Hope someone can help show why the equation ${\psi _0}(b + n) - {\psi _0}(b)=\frac{1}{b+n-1}+\frac{1}{b+n-2}+...+\frac{1}{b+1}+\frac{1}{b}$ holds. Thank you.

The restriction is that $n$ is positive integer and $b>0$.

• Which equation in particular would you like demonstrated? – Chappers Mar 31 '15 at 20:12
• @Chappers Thanks for your reply. I have refined the question. – Tony Mar 31 '15 at 20:14

Right, the definition of the Gamma function is $$\Gamma(z) = \int_0^{\infty} x^{z-1}e^{-x} \, dx,$$ for $z>0$. It is easy to show, using integration by parts, that $$\Gamma(z+1) = z \Gamma(z), \tag{1}$$ and $\Gamma(1)=1$ is also easy (it follows that $\Gamma(n)=(n-1)!$ for $n$ a positive integer, but I'm not going to need that.)
The digamma function (unhelpfully written $\psi(z)$) is defined as $$\psi(z) = \frac{\Gamma'(z)}{\Gamma(z)} = \frac{d}{dz} \log{\Gamma(z)}.$$ Now, taking logarithms of both sides of (1) and differentiating, we find that $$\log{\Gamma(z+1)} = \log{z}+\log{\Gamma(z)}\\ \psi(z+1) = \frac{1}{z} + \psi(z).$$ Therefore, $\psi(z+1)-\psi(z) = \frac{1}{z}$, and we can iterate this to get $$\psi(z+2)-\psi(z) = (\psi(z+2)-\psi(z+1))+(\psi(z+1)-\psi(z)) = \frac{1}{z+1} + \frac{1}{z},$$ and doing the same trick to $\psi(b+n)-\psi(b)$, we get to $$\psi(b+n)-\psi(b) = \frac{1}{b+n-1} + \dotsb + \frac{1}{b+1} + \frac{1}{b}.$$