I saw this formula and I don't know if it has a certain name or how can it be proven
If $|x|<1$ then $$1+2x+3x^2+4x^3+\cdots= \frac{1}{(1-x)^2}$$
Can someone tell me anything about it? How can it be proven?
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Diego Silvera
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Mihai Bujanca
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1What's $n$ doing there? – Git Gud Feb 17 '13 at 23:03
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1Definitely untrue for certain values of $n$ and $x$ – Brian B Feb 17 '13 at 23:03
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2You mean $1+2x+\ldots+nx^{n-1}+\ldots$. Just take the derivative of $1+x+\ldots+x^n+\ldots=\frac{1}{1-x}$ and justify term-by-term differentiation. – Julien Feb 17 '13 at 23:03
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You skipped over $3$, it goes $2x^2$ and $4x^3$. I'm not sure whether it's a typo or it should be $2n x^{n+1}$ or $2^nx^{n+1}$, or whatever. – Asaf Karagila Feb 17 '13 at 23:04
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@BujancaMihai Do you know that $\displaystyle \frac{1}{1-x}=1+x+x^2+\cdots$ for any $x\in \Bbb R$ such that $|x|<1$? – Git Gud Feb 17 '13 at 23:08
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2@BujancaMihai: See this Wikipedia page. – Zev Chonoles Feb 17 '13 at 23:10
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@ZevChonoles thanks a lot – Mihai Bujanca Feb 17 '13 at 23:13
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2Here is a similar question. – David Mitra Feb 17 '13 at 23:15
3 Answers
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This is a power series: $$\sum_{n=0}^\infty(n+1) x^n$$
For power series we can integrate term by term and we have:
$$\int\left(\sum_{n=0}^\infty(n+1) x^n\right)dx = \sum_{n=0}^\infty\int (n+1)x^ndx = \sum_{n=0}^\infty x^{n+1} = \sum_{n=1}^\infty x^n = \frac x{1-x}$$
So if we differentiate $\frac{x}{1-x}$ with respect to $x$, and we get the sum of the original series:
$$\left(\frac x{1-x}\right)'=\frac1{1-x}+\frac{x}{(1-x)^2}=\frac1{(1-x)^2}$$
As wanted. Of course we have to have $|x|<1$ for everything to converge and have meaning.
Asaf Karagila
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If you know that $$ 1+x+x^2+x^3 + \cdots = \frac{1}{1-x} $$ then you can get this identity by differentiating both sides.
This raises a problem: Is the derivative of the sum equal to the sum of the derivatives even when there are infinitely many terms? In fact, in some cases it is not. However, it can be shown that for a power series (i.e. each term is a number times a different power of $x$) that does work.
That, however, is not the only way to do it.
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1Can you please give me an example (link or whatever) in which when you have infinite terms the derivative is not the sum of derivatives? – MyUserIsThis Feb 17 '13 at 23:17
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1Let $g_n(x)=\dfrac{\sin((n+1)x)}{\sqrt{n+1}} - \dfrac{\sin(nx)}{\sqrt{n}}$. Then $\dfrac{d}{dx} \sum_{n=1}^\infty g_n(x)=0$, but $\sum_{n=1}^\infty g_n'(x)\ne 0$. (In particular the latter at $x=0$ diverges to $\infty$ rather than converging to $0$.) – Michael Hardy Feb 17 '13 at 23:47
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Here is a solution which doesn't use integration or differentiation of power series.
I am going to use $$1+x+x^2+x^3 + \cdots +x^k= \frac{1-x^{k+1}}{1-x}$$
is well known and easy to prove. I will assume that you know it, but if you don't I can provide a proof.
Let $$S_k=1+2x+3x^2+..+(k+1)x^k $$ $$T_k=1+x+x^2+..+x^k$$
Then
$$S_k-T_k=x+2x^2+3x^3+..+kx^k=xS_{k-1}=x(S_k-(k+1)x^k )$$
Thus
$$S_k(x-1)=(k+1)x^{k+1}-T_k=(k+1)x^{k+1}-\frac{1-x^{k+1}}{1-x}$$
Thus
$$S_k= \frac{1}{(1-x)^2}+\frac{(k+1)x^{k+1}}{x-1}-\frac{x^{k+1}}{(x-1)^2}$$
Since $\lim_k (k+1)x^{k+1}=\lim_k x^{k+1} =0 \, \forall |x|<1$ you get the desired result.
N. S.
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