Generally Geometric
Generalise the sum of a GP by using derivatives to make the
coefficients into powers of the natural numbers.
Age
16 to 18
Challenge level
The sum of the infinite geometric series $1 + x + x^2 + x^3 + \cdots$ and the binomial series are well known. How are the two related?
Show that $$\sum_{n=0}^\infty n x^n = {x\over(1-x)^2}$$ and find $$\sum_{n=0}^\infty n^2x^n.$$ Outline a method for finding $$\sum_{n=0}^\infty n^kx^n$$ where you do not have to carry out this computation beyond $k=2$.
Experiment with other expansions to try to find out the values for other interesting series.
Differentiate the geometric series term by term. Equate this to the
derivative of the sum of the series. The required identity follows
with a little simplification of the expressions to obtain terms
with the required coefficients.
Now carry out this process using the formula for the infinite sum.
Now carry out this process using the formula for the infinite sum.
Andy from Clitheroe Royal Grammar School sent us his work on this problem. He's given us two methods; can you see why he prefers the second one?
We begin by summing the series
$x+2x^2+3x^3+4x^4+\cdots$
| $x$ | + | $x^2$ | + | $x^3$ | + | $x^4$ | + | $\cdots$ |
| + | $x^2$ | + | $x^3$ | + | $x^4$ | + | $\cdots$ | |
| + | $x^3$ | + | $x^4$ | + | $\cdots$ | |||
| + | $\cdots$ |
In other words, we are writing it as a sum of geometric series!
Now, let us factorise the above sum as follows:
$(x + x^2 + x^3 + x^4+\ldots)(1 + x + x^2 + x^3 + x^4+\ldots)$
Wow, a product of geometric series!
We can then take a factor of $x$ out the first bracket to leave us with
$x(1 + x + x^2 + x^3+\ldots)^2$
Using the geometric sum given in the question, this comes to $$x\times \left(\frac{1}{1-x}\right)^2 = \frac{x}{(1-x)^2}$$ __
A similar method could be used for the series $x + 4x^2 + 9x^3 + 16x^4 +\ldots$, factorising it as $(x + 3x^2 + 5x^3 + 7x^4+\ldots)(1 + x + x^2 + x^3 +\ldots)$, then writing the left hand bracket as $(x + x^2 + x^3+\ldots + 2x^2 + 4x^3 + 6x^4+\ldots)$, from which point we can use our previous sum to obtain an answer. Unfortunately this doesn't generalise easily into higher powers, the amount of working needed growing much larger at each stage.
A more elegant solution is differentiation. If we differentiate our first series, we get $1 + 4x + 9x^2 + 16x^3+\ldots + n^2x^{n-1}+\ldots$. Multiplying through by $x$ gives us $x + 4x^2 + 9x^3+\ldots + n^2 x^n+\ldots$, which is the $n^2 x^n$ series we need.
If $x + 2x^2 + 3x^3 + 4x^4+\ldots = x/(1-x)^2$ then $x(d[x + 2x^2 + 3x^3+\ldots]/dx) = x(d[x/(1-x)^2]/dx)$.
But the left-hand side is equal to $x + 4x^2 + 9x^3 + 16x^4 +\ldots$, the sequence we want to sum.
We can resolve the right-hand using the quotient rule, and it comes to $x(1+x)/(1-x)^3$.
__
To take it into higher powers, notice that
$d[x + 4x^2 + 9x^3+\ldots]/dx = 1 + 8x + 27x^2+\ldots$.
Therefore $x d[x + 4x^2 + 9x^3+\ldots]/dx = x + 8x^2 + 27x^3+\ldots$, our next sequence. We can differentiate the previous infinite sum and multiply by $x$ at each stage to get the sum for the next power, and by applying the same process to the closed-form expression, we can obtain a closed-form expression for the next power.
Differentiating the known sum, then differentiating the series term
by term and multiplying by $x$ gives the first result. I f this
method works once it will work twice or $k$ times.