Sum to infinity of a Series

For $|x|<1$ we have $$P=x(1+2x+3x^2+...+)=x(x+x^2+x^3+...)'=x\cdot\left(\frac{x}{1-x}\right)'=...$$ Can you end it now?

You can find where the series converges using D'Alembert's convergence test. Namely,

$$\lim\limits_{n\to\infty}\frac{|(n+1)x_{n+1}|}{|nx_n|} = \lim\limits_{n\to\infty}\left| \frac{n+1}{n}\right||x| = |x|$$ If this limit is less than 1 (i.e. $|x|< 1$), then the series converges. When $|x|>1$, then the series diverges. For $|x|=1$, the test is inconclusive, but we can verify by plugging in $x=1$ and $x=-1$ respectively, that those series diverge. Another special case to consider is $x=0$, for which the series obviously converges (and its sum is 0).

Thus, we conclude that this series converges iff $|x|<1$.

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Now I will address your question about the permutation of elements.

Let $$P_N=\sum\limits_{n=1}^Nnx^n$$ Note that $$P=\lim\limits_{N\to\infty}P_N$$

We have $$P_N=x\cdot \sum\limits_{n=1}^N nx^{n-1} = x \cdot \left(\underbrace{1+x+...+x^{N-1}}+\underbrace{x+2x^2+...+(N-1)x^{N-1}}\right) =$$ $$= x \cdot \left( \sum\limits_{n=0}^{N-1} x^n + \sum\limits_{n=1}^{N-1} nx^n\right) = x \cdot \left( \frac{1-x^N}{1-x} + P_{N-1} \right)$$

Letting $N\to\infty$, we arrive at the same result as you did. You can also notice that I did the same manipulations as you did, the only difference is that I was considering finite sums.

There is a theorem that states that if a series converges absolutely (the one in your example does), then any series obtained by any permutation of the summands converges and it has the same sum as the starting series.

If you find yourself unsure whether you can permute the summands, try to consider finite sums, because then you can use any manipulations that apply to finite sums, including permutation of elements. After that, just take the limit as $N \to \infty$.

Hint: For the finite $$\sum_{i=1}^{n}ix^i$$ we get $$\sum_{i=1}^{n}i x^i=\frac{(n x-n-1) x^{n+1}+x}{(1-x)^2}$$