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Problem: Find the sum of the series $$x+2x^2+3x^3+4x^4+... = \sum\limits_{n=1}^\infty nx^n$$

My solution is as follows: Let P be the above series. $$P=x(1+2x+3x^2+4x^3+...) =x(1+x+x^2+x^3+x^4+...x+2x^2+3x^3+4x^4) =x\left(\frac{1}{1-x}+P\right)$$

Then by some trivial algebraic manipulation we obtain $$P=\frac{x}{(1-x)^2}$$

However, I am mainly concerned about a technical issue in my solution. From the first to the second step, I sort of "rearranged" /“split” the terms in my series so as to massage it into a desired form. Yet I am unable to prove that the series in the bracket is convergent or even absolutely convergent, so I am scared that a permutation of the terms might affect the limit that the series actually approaches. (I have heard of the Riemman Series Theorem before). Is there thus any way I can prove the convergence of the series in the brackets? I may even be going in circles and have confused myself in the process.

Any help will be much appreciated. Thanks in advance.

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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 $|x|<1$.


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 $N \to \infty$.

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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?

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

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