Find the expression for the sum of this power series

[Kqwert]

Thank you, but what cancellation? Re-writing (n-1)/(n-1)! to 1/(n-2)! ? Not sure if I understand exactly what happens and why.

 

[Orodruin]

Thank you, but what cancellation? Re-writing (n-1)/(n-1)! to 1/(n-2)! ? Not sure if I understand exactly what happens and why.

Yes, that cancellation. Because, as I said in #24, for $n = 1$, that term reads $0/0!$ and the terms you would cancel are 0 and 0, which you cannot do. You have to look at that term separately and realize that it is actually zero so that it gives no contribution.

 

[Kqwert]

Yes, that cancellation. Because, as I said in #24, for $n = 1$, that term reads $0/0!$ and the terms you would cancel are 0 and 0, which you cannot do. You have to look at that term separately and realize that it is actually zero so that it gives no contribution.

I understand that, but I don't understand the order of things. When we start from

Should we first say that

and then do the change of variables? I am becoming a bit confused re. the best way of solving this.

 

[Orodruin]

What you had from the beginning was
$$
\sum_{n=1}^\infty \frac{(n-1)x^n}{(n-1)!},
$$
not the sum from $n = 0$. It does not matter if you change variables first or if you do the cancellation first. The only thing you must do is to remove the term $n = 1$ (or equivalently $q = -1$) before you do the cancellation because that step is not valid for that term.

 

[Kqwert]

And by removing the n = 1 term we end up starting at n = 2, right?

 

[Orodruin]

Yes, which means $q = 0$.

 

[Kqwert]

Excellent, thanks a lot for the help!

 

[Orodruin]

So now that you have worked through it, let me just mention an alternative approach, which is how I initially did it but it involves some extra trickery using derivatives. The aim is to reduce the remaining series to that of the exponential function by extracting powers of $x$ and using a derivative relation.

The point is to note that $n x^{n-1}$ is the derivative of $x^n$. We have
$$
\sum_{n=1}^\infty \frac{nx^n}{(n-1)!} = x \sum_{n=1}^\infty \frac{n x^{n-1}}{(n-1)!} = x \frac{d}{dx} \sum_{n=1}^\infty \frac{x^{n}}{(n-1)!}
$$
where in the first step we have just factorised out an $x$ and in the second we applied the derivative relation. Now we can extract another factor $x$ from the sum and end up with
$$
\sum_{n=1}^\infty \frac{nx^n}{(n-1)!} = x\frac{d}{dx}\left(x \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}\right)
= x\frac{d}{dx}\left(x \sum_{m=0}^\infty \frac{x^{m}}{m!}\right) = x \frac{d}{dx}(x e^x) = x (x e^x + e^x) = x(1+x) e^x,
$$
where in the middle we did the substitution $m = n-1$ then used the power series for $e^x$ and regular differentiation. Luckily, this is the same result as you got.

 

[Ray Vickson]

Excellent, thanks a lot for the help!

I notice that you were getting all mixed up with summations (what terms should be removed, how should the sums be re-indexed, etc?). When I solve such problems I try to avoid all that by writing out a few terms explicitly:
$$S = \sum_{n=1}^\infty \frac{n x^n}{(n-1)!} = x + \frac{2 x^2}{1!}+\frac{3 x^3}{2!} + \frac{4 x^4}{3!} + \cdots + \frac{n x^n}{(n-1)!} + \cdots $$ This can be written as
$$S = x + \frac{(1+1) x^2}{1!} + \frac{(2+1) x^3}{2!} + \frac{(3+1)x^4}{3!} + \cdots + \frac{((n-1)+1) x^n}{(n-1)!)} + \cdots, $$ so $S = S_1 + S_2,$ where
$$S_1 = x^2 + \frac{2 x^3}{2!} + \frac{3 x^4}{3!} +\cdots + \frac{(n-1) x^{n-1}}{(n-1)!} +\cdots\\ = x^2 \left(1 + x + \frac{x^2}{2!} + \cdots + \frac{x^{(n-2)}}{(n-2)!} + \cdots \right), $$ and
$$S_2 = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \cdots + \frac{x^n}{(n-1)!} + \cdots \\= x \left (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots+ \frac{x^{n-1}}{(n-1)!} + \cdots \right).$$ The rest is easy now.

When I say that I, personally, write out a few terms (as above) I really mean it; I have found over the years that doing that eliminates a lot of confusion and is a good way of avoiding a headache.