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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, 11 months ago

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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


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.

Cody Johnson - 3 years, 11 months ago

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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.



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

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

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



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


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

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

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 \(2^3 \cdot \left( \frac{1}{4} + \frac{1}{2} \right) = \boxed{6}\).

Ivan Koswara - 3 years, 11 months ago

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