Question

Show that
\[\left(\sum_{k=0}^\infty x^k\right)^2=\sum_{k=0}^\infty (k+1)x^{k}\]
I encourage anyone to try to find a "closed" form for
\[\left(\sum_{k=0}^\infty x^k\right)^n\]
so that a sum raised to a power is equal to another sum not raised to a power. \((n\in\mathbb{N})\)

Answer 1

Hmm, I think ultimately it would involve integration.

Answer 2

The right side would only converge if \(x<1\), I believe. It's a geometric series.

Answer 3

Right, let's also assume \(|x|<1\) for convenience!

Answer 4

At first I had a tedious (sort of combinatoric) approach in mind that involves expanding the polynomial \((1+x+x^2+\cdots)^2\) and I noticed the pattern above, so I was wondering if there's one for any natural power \(n\).

We could also do this by recognizing that the LHS is the power series for \(\dfrac{1}{(1-x)^2}\) and integrating/differentiating the RHS to establish equality, which seems to be the direction you're going in.

Answer 5

\[
\left(\sum_{k=0}^\infty x^k\right)^n
= \left(\frac{1}{1-x}\right)^n
=(1-x)^{-n}
\]

Answer 6

For \(n=3\) we get
\[\left(\sum_{k=0}^\infty x^k\right)^3=\sum_{k=0}^\infty \frac{(k+1)(k+2)}{2}x^k\]
and for \(n=4\),
\[\left(\sum_{k=0}^\infty x^k\right)^3=\sum_{k=0}^\infty \frac{(k+1)(k+2)(2k+3)}{6}x^k\]
with those coefficients representing triangular and tetrahedral numbers,
\[\sum_{r=1}^{k+1}r^2=\frac{(k+1)(k+2)}{2}\quad\text{and}\quad\sum_{r=1}^{k+1}=\frac{(k+1)(k+2)(2k+3)}{6}\]
so there might not be a "closed" form for any \(n\)...

Answer 7

Not that I'm aware of, at any rate.

Answer 8

Oops,\[\sum_{r=1}^{k+1}\color{red}r=\frac{(k+1)(k+2)}{2}\quad\text{and}\quad\sum_{r=1}^{k+1}\color{red}{r^2}=\frac{(k+1)(k+2)(2k+3)}{6}\]

Answer 9

Hmm, is the binomial theorem valid for negative powers?

Answer 10

We can say this much I think:
\[\left(\sum_{k=0}^\infty x^k\right)^n=\sum_{k=0}^\infty \left(\sum_{r=1}^{k+1}r^{n-2}\right)x^k\]

Answer 11

I'm taking Maclaurin series for \((1-x)^{-n}\)

Answer 12

\[
f^{(0)}(0) = 1\\
f^{(1)}(0) = n\\
f^{(2)}(0) = (n+1)n\\
f^{(3)}(0) = (n+2)(n+1)n\\
f^{(k)}(0) = (n+k-1)\ldots (n+1)n = \frac{(n+k-1)!}{(n-1)!}\\
\]

Answer 13

I think this means: \[
(1-x)^{-n}=\sum_{k=0}^{\infty }\frac{(n+k-1)!}{(n-1)!} \frac{x^k}{k!}
\]

Answer 14

But that would mean: \[
(1-x)^{-n}=\sum_{k=0}^{\infty }\frac{(n+k-1)!}{(n-1)!} \frac{x^k}{k!} = \sum_{k=0}^{\infty }{n+k-1\choose k}x^k
\]

Answer 15

Which would make your equality seem a bit strange.

Answer 16

Nice, the formula seems to check out in Mathematica

Answer 17

For \(n=2\), we get
\[\sum_{k=0}^\infty \binom {k+1}kx^k=\sum_{k=0}^\infty (k+1)x^k\]