Closed form solution method for infinite sum (but not "by inspection")

I have a series whose terms are defined by the recurrence relation

$$T_n = \alpha \ \left(1+\frac{\mu}{n}\right)\ T_{n-1}$$ with $T_0=1$

So $$T_n=\alpha^n\prod_{k=1}^n \left(1+\frac{\mu}{k}\right)\tag{1}\label{eq1}$$

$\alpha$ and $\mu$ are real constants with $0\le\alpha\lt1$ and $\mu\ge 0$. I want the sum $$S(\alpha,\mu)=\sum_{n=0}^\infty T_n$$ Now, I know that when $\mu=0$, $$S(\alpha,0)=\frac{1}{1-\alpha}$$

Here's the funny thing... I was playing around in Excel with actual numbers and stumbled upon the general solution:$$S(\alpha,\mu)=\left(\frac{1}{1-\alpha}\right)^{1+\mu}$$

Does anyone know any general techniques that would have led me to this expression?

Here's another funny thing... if I expand the solution in a Taylor series around $\alpha=0$, I get $$S(\alpha,\mu)=1+(1+\mu)\alpha+\frac12(1+\mu)(2+\mu)\alpha^2+\frac16(1+\mu)(2+\mu)(3+\mu)\alpha^3+\ldots$$ Each term in the Taylor series is $$\frac{\alpha^n}{n!}\prod_{k=1}^n(k+\mu)=\alpha^n\prod_{k=1}^n\frac{k+\mu}{k} =\alpha^n\prod_{k=1}^n\left(1+\frac{\mu}{k}\right)$$ which is of course $T_n$ as defined in $\eqref{eq1}$

I suppose if I were really good, I might have recognized the Taylor series expansion in the first place, but I'm not that good. :) So does anyone know a better method by which I could have arrived at the answer?

Welcome to MSE!

It sounds like you'd probably be interested in generating functions! Indeed, you're well on your way to rediscovering them yourself ^_^

The idea is to take a recurrence and turn it into a power series, then do some manipulations to this power series in order to learn things about your recurrence. There's a whole book dedicated to general techniques for computing generating functions (without needing to be clever) called generatingfunctionology. I really can't recommend it highly enough!

Now, a generating function is an expression of the form

$$\sum a_n x^n$$

and we can see that your sum

$$S(\mu,\alpha) \triangleq \sum \left ( \prod_{k=1}^n \left ( 1 + \frac{\mu}{k} \right ) \right ) \alpha^n$$

is of this form (where we're using the variable $\alpha$ instead of $x$ to be consistent with your question).

Now we want to find a "closed form" for this power series. That is, we want to find a nice (ideally rational) function which gives us $S$ when we taylor expand at $\alpha$. Viewed through this lens your observation about the taylor series is kind of a tautology, but I'll reemphasize that the idea to compute a closed form of a power series using calculus is already a good one!

Let's work through deriving a closed form together. At some points it might look like I'm being clever, but that's only because this answer is too short to explain all the tricks of the trade. I assure you, though, if you read the book I linked earlier (and in particular if you work through the exercises) you'll become fluent enough to do this all yourself too!

Let's write $s_n = \prod_{k=1}^n \left ( 1 + \frac{\mu}{k} \right )$ so that our sum becomes $S(\mu,\alpha) = \sum s_n \alpha^n$. Then we get the new recurrence relation

$$ s_{n+1} = \left ( 1 + \frac{\mu}{n+1} \right ) s_n = s_n + \frac{\mu}{n+1} s_n.$$

Now we do what we always do with generating functions: Multiply both sides by $\alpha^n$

$$ s_{n+1} \alpha^n = s_n \alpha^n + \frac{\mu}{n+1} s_n \alpha^n $$

then sum over all $\alpha^n$

$$\sum s_{n+1} \alpha^n = \sum s_n \alpha^n + \sum \frac{\mu}{n+1} s_n \alpha^n.$$

If we remember that $S = \sum s_n \alpha^n$, we can rewrite this as

$$\frac{S - s_0}{\alpha} = \frac{S - 1}{\alpha} = S + \mu \sum \frac{s_n}{n+1} \alpha^n.$$

Now it's time for the trick which may look clever at first, but is really part of the general theory. We see the power series $\sum \frac{s_n}{n+1} \alpha^n$ and we ask ourselves: "how do we usually get $n+1$s in the denominator?" The answer, of course, is "after integrating!" Indeed

$$ \int S \ d\alpha = \int \sum s_n \alpha^n \ d\alpha = \sum \int s_n \alpha^n \ d\alpha = \sum \frac{s_n}{n+1} \alpha^{n+1} $$

which looks familiar!

Putting this into our earlier expression gives

$$\frac{S - 1}{\alpha} = S + \frac{\mu}{\alpha} \int S\ d\alpha$$

which we want to solve for $S$. We can already make some progress by turning this into

$$(1 - \alpha)S = 1 + \mu \int S\ d\alpha$$

but this looks a bit tricky. I don't know about you, but I'm much more comfortable with differential equations than I am with integral equations. So let's differentiate both sides with respect to $\alpha$ to get

$$-S + (1-\alpha) \frac{dS}{d\alpha} = \mu S$$

thankfully this is separable, and we solve (using the initial condition $S(0) = s_0 = 1$ to see

$$S = \left ( \frac{1}{1-\alpha} \right )^{\mu + 1}$$

which agrees with what you computed in excel!

Hopefully seeing this worked out in full gives you some proof that there really is a methodology here, and hopefully the rabbits that I did have to pull from a hat felt like tricks you yourself could come to learn.

I hope this helps ^_^

We can rewrite $ T_{n} $ as follows : $$ T_{n}=\frac{\alpha^{n}}{n!}\prod_{k=1}^{n}{\left(k+\mu\right)}=\frac{\left(-\alpha\right)^{n}}{n!}\prod_{k=0}^{n-1}{\left(-1-\mu-k\right)}=\left(-\alpha\right)^{n}\binom{-1-\mu}{n} $$

Using $ \sum\limits_{n=0}^{+\infty}{\binom{\alpha}{n}x^{n}}=\left(1+x\right)^{\alpha} $, we get that $ \sum\limits_{n=0}^{+\infty}{T_{n}}=\left(1-\alpha\right)^{-1-\mu} $.