# Interesting question - Not homework

matt grime
Homework Helper
what's wrong with that?

Nothing! It has just dawned on me that because I copy and paste TEX through my working that the 2^(2n) was more than likely a mistype otherwise I would not have arrived at my solution!!

HallsofIvy
Homework Helper
A couple of hours after my post (when I was not near a computer!) it suddenly dawned on me that NewScientist must have meant
$$\Sigma log_{2^{2^n}} e$$

That reduces to
$$\Sigma \frac{(-1)^n}{ln(2^{2^n}}= \Sigma \frac{(-1)^n}{2^n ln 2}= \frac{1}{ln 2}\Sigma \left(\frac{-1}{2}\right)^n$$
which is, in fact, a simple geometric series.

benorin
Homework Helper
Gold Member
That's Beautiful.

HallsofIvy said:
A couple of hours after my post (when I was not near a computer!) it suddenly dawned on me that NewScientist must have meant

$$\Sigma log_{2^{2^n}} e$$

That reduces to

$$\Sigma \frac{(-1)^n}{ln 2^{2^n} }= \Sigma \frac{(-1)^n}{2^n ln 2}= \frac{1}{ln 2}\Sigma \left(-\frac{1}{2}\right) ^n$$
which is, in fact, a simple geometric series.

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