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# Square over power of two sum

What is the value of $\sum_{ i=1 }^n \frac{i^2}{2^i}$ when n approaches infinity?

Note by Takeda Shigenori
3 years, 1 month ago

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Let

$S=\sum_{i=1}^\infty\frac{i^2}{2^i}$

We have

\begin{align*} 2S&=\sum_{i=1}^\infty\frac{i^2}{2^{i+1}}\\&=\sum_{i=0}^\infty\frac{(i+1)^2}{2^i}\\&=1+\sum_{i=1}^\infty\frac{(i+1)^2}{2^i} \end{align*}

Subtracting the two, we get

\begin{align*} 2S-S&=S\\&=1+\sum_{i=1}^\infty\frac{(i+1)^2-i^2}{2^i}\\&=1+\sum_{i=1}^\infty\frac{2i+1}{2^i}\\&=1+\sum_{i=1}^\infty\frac{2i}{2^i}+\sum_{i=1}^\infty\frac{1}{2^i}\\&=2+2\sum_{i=1}^\infty\frac{i}{2^i} \end{align*}

Now, this is an arithmetico-geometric series. Hence, its sum is

$S=2+\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{\frac{1}{2}\cdot1}{\left(1-\frac{1}{2}\right)^2}=\boxed{6}$

Now instead, let's pretend that this is the day of the test and you forgot the formula for a arithmetico-geometric series. What would you do? Well, it appears as though our manipulation turned the numerator from $$i^2$$ to $$i$$. Hm. Maybe if we manipulate it more, we can turn that $$i$$ to a $$1$$ and then solve using geometric series! Let's see:

$S'=\sum_{i=1}^\infty\frac{i}{2^i}$ \begin{align*} 2S'&=\sum_{i=1}^\infty\frac{i}{2^{i+1}}\\&=\sum_{i=0}^\infty\frac{i+1}{2^i}\\&=1+\sum_{i=1}^\infty\frac{i+1}{2^i} \end{align*} $2S'-S'=S'=1+\sum_{i=1}^\infty\frac{1}{2^i}$

Hooray! It worked! So $$S'=1+1=2$$ from geometric series, hence $$S=2+2S'=2+2(2)=\boxed{6}$$, again. · 3 years, 1 month ago

Lemma 1: $$\displaystyle\sum_{i=k}^\infty \dfrac{1}{2^i} = \dfrac{1}{2^{k-1}}$$. This is just geometric series.

Express $$i^2 = 1 + 3 + 5 + \ldots + (2i-1)$$ and collect terms with equal numerators. We then have the sum to be:

$$\left( \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \ldots \right) + \left( \dfrac{3}{2^2} + \dfrac{3}{2^3} + \ldots \right) + \left(\dfrac{5}{2^3} + \ldots \right) + \ldots$$

$$= \displaystyle\sum_{j=1}^\infty \sum_{i=j}^\infty \dfrac{2j-1}{2^i}$$

$$= \displaystyle\sum_{j=1}^\infty (2j-1) \cdot \dfrac{1}{2^{j-1}}$$ by Lemma 1.

Express $$2j-1 = 1 + 2 + 2 + \ldots + 2$$ and collect terms with equal numerators, considering the $$2$$s as distinct (so we collect the terms caused by the first $$2$$, then caused by the second $$2$$, and so on). The sum becomes:

$$\left( \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \ldots \right) + \left( \dfrac{2}{2^2} + \dfrac{2}{2^3} + \ldots \right) + \left(\dfrac{2}{2^3} + \ldots \right) + \ldots$$

$$= \displaystyle\sum_{j=0}^\infty \dfrac{1}{2^j} + \sum_{j=1}^\infty \sum_{i=j}^\infty \dfrac{2}{2^i}$$

$$= \dfrac{1}{2^{-1}} + \displaystyle\sum_{j=1}^\infty 2 \cdot \dfrac{1}{2^{j-1}}$$ by Lemma 1.

$$= 2 + 2 \cdot \dfrac{1}{2^{-1}} = \boxed{6}$$

Another solution:

Lemma 1: $$(1+x+x^2+x^3+\ldots)^3 = \sum_{i=0}^\infty \binom{i+2}{2} x^i$$. This is from the generating function solution of solving $$a+b+c=i$$ where $$a,b,c$$ are nonnegative integers.

Thus,

$$(1+x+x^2+x^3+\ldots)^3x$$

$$= \displaystyle\sum_{i=0}^\infty \binom{i+2}{2} x^{i+1}$$

$$= \displaystyle\sum_{i=1}^\infty \binom{i+1}{2} x^i$$

$$= \displaystyle\sum_{i=1}^\infty \dfrac{i(i+1)}{2} x^i$$

and

$$(1+x+x^2+x^3+\ldots)^3x^2$$

$$= \displaystyle\sum_{i=1}^\infty \dfrac{i(i+1)}{2} x^{i+1}$$ from previous result

$$= \displaystyle\sum_{i=2}^\infty \dfrac{(i-1)i}{2} x^i$$

$$= \displaystyle\sum_{i=1}^\infty \dfrac{i(i-1)}{2} x^i$$ because at $$i=1$$ the term gives $$0$$

So we have

$$(1+x+x^2+x^3+\ldots)^3(x^2+x)$$

$$= \displaystyle\sum_{i=1}^\infty \left( \dfrac{i(i-1)}{2} + \dfrac{i(i+1)}{2} \right) x^i$$

$$= \displaystyle\sum_{i=1}^\infty i^2 x^i$$

Substituting $$x = \dfrac{1}{2}$$, the right hand side gives the expression we need, while the left hand side can now be easily evaluated to $$2^3 \cdot \left( \frac{1}{4} + \frac{1}{2} \right) = \boxed{6}$$. · 3 years, 1 month ago