Exercise 6.5.6

Question

Previous work on geometric series (Example 2.7.5) justifies the formula
$$
\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots, \quad 	ext { for all }|x|<1 .
$$
Use the results about power series proved in this section to find values for $\sum_{n=1}^{\infty} n / 2^{n}$ and $\sum_{n=1}^{\infty} n^{2} / 2^{n}$. The discussion in Section 6.1 may be helpful.

Ulisse Mini · Accepted Answer

Let $a_n = 1$; we have
    $$\sum^\infty_{n=0}a_n x^n = \frac{1}{1-x}$$
    with a radius of convergence of 1. By Theorem 6.5.6 we can differentiate this termwise, to get
    $$\sum^\infty_{n=1} n a_n x^{n-1} = \sum^\infty_{n=0} (n+1) x^n = \sum^\infty_{n=0} x^n + \sum^\infty_{n=1}nx^n = \frac{1}{1-x} + \sum^\infty_{n=1}nx^n = \frac{1}{(1-x)^2}$$
    $$\sum^\infty_{n=1} nx^n= \frac{x}{(1-x)^2}$$
    Substituting $x=1/2$ we have $\sum^\infty_{n=1}n/2^n = 2$. We can differentiate the series again to get
    $$\sum^\infty_{n=1}(n^2 + n) x^{n-1} = \sum^\infty_{n=0} n^2x^n + 3\sum^\infty_{n=0}nx^n + 2\sum^\infty_{n=0} x^n = \frac{2}{(1-x)^3}$$
    Substituting $x =1/2$ we have $\sum^\infty_{n=1}n^2/2^n = 6$.

Exercise 6.5.6

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Exercise 6.5.6

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