3

Find a "closed form" of summation $n=2$ to infinity $n(n-1)x^n$.

I don't have much clue to solve this. Can anyone please explain how to approach this?

gegu
  • 1,694
rohit
  • 371

2 Answers2

4

Just rephrasing what André Nicolas wrote, consider the term $$u_n=n(n-1)x^n=x^2\Big(n(n-1)x^{n-2}\Big)$$ I am sure that you recognize that what is inside the parentheses is the second derivative of $x^n$. So, $$\sum_{n=2}^{\infty} u_n=x^2\frac{d^2}{dx^2}\Big(\sum_ {n=2}^{\infty} x^n\Big)$$ and in the rhs, you have a geometric progression; so, provided a suitable $x$, $$\sum_ {n=2}^{\infty} x^n=\frac{x^2}{1-x}$$ Taking derivatives and simplifying leads to $$\frac{d^2}{dx^2}\Big(\sum_ {n=2}^{\infty} x^n\Big)=\frac{2}{(1-x)^3}$$ and, finally,$$\sum_{n=2}^{\infty} u_n=\frac{2 x^2}{(1-x)^3}$$

Claude Leibovici
  • 260,315
2

A start: We have for suitable $x$ $$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots.$$ Differentiate twice.

André Nicolas
  • 507,029
  • i got u why we are doing this but not sure of how to put that in proper form... – rohit Oct 21 '14 at 04:52
  • If you want to justify why term by term differentiation is correct for $|x|\lt 1$, you have to appeal to a general theorem. We are inside the interval of convergence of a power series, so term by term differentiation of the right-hand side yields the same result as differentiation of the left-hand side. You know of course how to differentiate the left-hand side twice. When you differentiate the right-hand side term by term, you will not quite get the correct powers of $x$, but multiplication by $x^2$ will fix that. – André Nicolas Oct 21 '14 at 04:59
  • I just saw your comment under your question. Differentiating the left-hand side twice gives $\frac{2}{(1-x)^3}$. Multiply by $x^2$ and you will have your answer. No summation. – André Nicolas Oct 21 '14 at 05:02
  • i think i got u. differentiation of 1/(1-x) twice is -1/(1-x)^3 multiplied by x^2. so the final answer to this will be -x^2/(1-x)^3...am i right? – rohit Oct 21 '14 at 05:02
  • ok got u...thnk u... – rohit Oct 21 '14 at 05:04
  • You got the wrong derivatives. When you differentiate $\frac{1}{1-x}$ you should get $\frac{1}{(1-x)^2}$ (Chain Rule). Doing it again gives $\frac{2}{(1-x)^3}$. – André Nicolas Oct 21 '14 at 05:04