Find the expression for the sum of this power series

In summary: But why does that matter when the expression has been re-written? isn't is possible to solve it from the re-written...Yes, it is possible to solve it from the re-written expression, but the q term will not cancel and you will get x^2*e^x.Yes, it is possible to solve it from the re-written expression, but the q term will not cancel and you will get x^2*e^x.
  • #36
Yes, which means ##q = 0##.
 
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  • #37
Excellent, thanks a lot for the help!
 
  • #38
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.
 
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  • #39
Kqwert said:
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.
 
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