Mildly Infuriating

Let $S = \displaystyle \sum_{n=1}^\infty \frac{n^{2}}{2^{n}}$ . Then:

$S = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + \ldots + \frac{k^{2}}{2^{k}} + \ldots$

Multiply both sides by two:

$2S = \frac{1^{2}}{2^{0}} + \frac{2^{2}}{2^{1}} + \frac{3^{2}}{2^{2}} + \ldots + \frac{(k + 1)^{2}}{2^{k}} + \ldots$

Subtracting $S$ from $2S$ :

$2S - S = 1 + \frac{3}{2} + \frac{5}{2^{2}} + \frac{7}{2^{3}} + \ldots + \frac{2k + 1}{2^{k}} + \ldots$

Then: $S = \displaystyle \sum_{i=0}^\infty \frac{2i + 1}{2^{i}}$

I will break this sum into two smaller ones: $S_{1}$ and $S_{2}$ , such that:

$\begin{cases} S_{1} = \displaystyle \sum_{i=0}^\infty \frac{2i}{2^{i}} \\ S_{2} = \displaystyle \sum_{i=0}^\infty \frac{1}{2^{i}} \end{cases}$

First, notice that $S_{2}$ is an infinite geometric series with first term equal to 1 and ratio $\frac{1}{2}$ ; thus, its value equals $\frac{1}{1 - \frac{1}{2}} = 2$ .

Now, let's work with $S_{1}$ .

$S_{1} = \displaystyle \sum_{i=0}^\infty \frac{2i}{2^{i}} = \displaystyle \sum_{i=0}^\infty \frac{i}{2^{i - 1}}$

$S_{1} = \frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \ldots + \frac{p}{2^{p - 1}} + \ldots$

Multiplying both sides by two:

$2S_{1} = 2 + \frac{2}{2^{0}} + \frac{3}{2^{1}} + \ldots + \frac{p + 1}{2^{p - 1}} + \ldots$

Subtracting $S_{1}$ from $2S_{1}$ :

$2S_{1} - S_{1} = 2 + 1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \ldots + \frac{1}{2^{p - 1}} + \ldots = 2 + \displaystyle \sum_{i=0}^\infty \frac{1}{2^{i }} = 2 + S_{2}$

Thus, $S_{1} = 2 + S_{2} = 2 + 2 = 4$ , and therefore $S = S_{1} + S_{2} = 2 + 4 = 6$ .

$\sum_{n=0}^{ \infty } x^{n} = \dfrac{1}{1-x} \quad \forall \quad |x| \leq 1$ $\Rightarrow \sum_{n=0}^{ \infty } ( \frac{d}{dx} x^{n} ) = \frac{d}{dx} ( \dfrac{1}{1-x} ) \\ \Rightarrow \sum_{n=0}^{ \infty} ( n \cdot x^{n} ) = \dfrac{x}{ (1-x)^{2} } \\ \Rightarrow \sum_{n=0}^{ \infty} \frac{d}{dx} (n \cdot x^{n} ) = \frac{d}{dx} ( \dfrac{x}{ (1-x)^{2} } ) \\ \Rightarrow \sum_{n=0}^{ \infty} ( n^{2} \cdot x^{n} ) = x\cdot \dfrac{ (1-x)^{2} + 2x\cdot (1-x) }{ (1-x)^{4} } \\= \frac{x+x^{2}}{ (1-x)^{3} }$

Well, we just input $x= \dfrac{1}{2}$ and we get the answer as $6$ .

$\sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } =\sum _{ n=0 }^{ \infty }{ \frac { { (n+1) }^{ 2 } }{ { 2 }^{ n+1 } } } \\ \qquad \qquad =\frac { 1 }{ 2 } \left( \sum _{ n=0 }^{ \infty }{ \left( \frac { { n }^{ 2 } }{ { 2 }^{ n } } +2\frac { { n } }{ { 2 }^{ n } } +\frac { 1 }{ { 2 }^{ n } } \right) } \right) \\ \qquad \qquad =\frac { 1 }{ 2 } \left( \sum _{ n=0 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } +2\sum _{ n=0 }^{ \infty }{ \frac { { n } }{ { 2 }^{ n } } } +\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ n } } } \right) \\ \qquad \qquad =\frac { 1 }{ 2 } \left( \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } +0+2(2)+2 \right) \\ \qquad \qquad =\frac { 1 }{ 2 } \left( \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } +6 \right) \\ \qquad \qquad =\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } +3\\ \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } -\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } =3\\ \\ \frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } =3\\ \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 2 }^{ n } } } =6\\$

Python:

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from fractions import Fraction as frac
n = 1
infinity = 1000
total = 0
while n <= infinity:
    total += frac(n*n, 2**n)
    n += 1
print "Answer:", float(total)