Root

you could try to use the upper bound of the sum through an integral. But this use some specials functions (the "exponential integral") and is not a very neat proof (IMO). Not that I have anything better to offer...

We have $\sum_{m=2}^\infty \frac{\ln m}{2^{m-1}} \; = \; -2\left(\frac{\partial}{\partial u} \mathrm{Li}_u(z)\right)\Big|_{u=0,z=\frac12} \; = \; 1.0156678 \; < \; 1.0986123 \; = \; \ln3$ using a derivative of the polylogarithm function.

Of course, the original series converges pretty rapidly, and so we could quickly show that its sum was less than $3$ without evaluating it explicitly. For example, $k < 1.02^k$ for $k \ge 50$ , and $\sum_{k \ge 50} \frac{\ln k}{2^{k-1}} \; < \; \ln1.02 \times \sum_{k \ge 50} \frac{k}{2^{k-1}} \; < \; 3.6 \times 10^{-15}$ and we can calculate $\sum_{k=2}^{49} \frac{\ln k}{2^{k-1}}$ readily enough and see that it is smaller than $\ln3 - 3.6 \times 10^{-15}$ .

I don't know if anyone is still looking at these old problems, but I just wanted to add a solution that involves only simple hand calculations.

My start was very similar to Mark's and I got the same infinite sum.

$\qquad\qquad\qquad\qquad S = \sum_{k=2}^{\infty} \frac{ln(k)}{2^{k-1}}\qquad\quad(Equation \quad 1)$

Notice that a term for $k=1$ can be added since that term will be $\frac{ln(1)}{2^{0}} = 0$ .

$S = \sum_{k=1}^{\infty} \frac{ln(k)}{2^{k-1}}$

Substitute $k-1$ for $k$ in this last equation so that this equation for $S$ sums from $k=2$ to $\infty$ the same way $Equation \quad 1$ does.

$S = \sum_{k=2}^{\infty} \frac{ln(k-1)}{2^{k-2}}$

Now the indices match between this equation and that of $Equation \quad 1$ , but the denominators corresponding to the same index do not. Pull out a factor of 2 from every element in this series so that the denomitators are the same in the two series.

$S = 2\sum_{k=2}^{\infty} \frac{ln(k-1)}{2 \cdot 2^{k-2}}$

$\qquad\qquad\qquad\qquad \frac{1}{2}S = \sum_{k=2}^{\infty} \frac{ln(k-1)}{2^{k-1}} \qquad\quad (Equation \quad 2)$

Subtract $Equation \quad 2$ from $Equation \quad 1$ :

$S - \frac{1}{2}S = \sum_{k=2}^{\infty} \frac{ln(k)}{2^{k-1}} - \sum_{k=2}^{\infty} \frac{ln(k-1)}{2^{k-1}}$

$\frac{1}{2}S = \sum_{k=2}^{\infty} \bigg[\frac{ln(k)}{2^{k-1}} - \frac{ln(k-1)}{2^{k-1}}\bigg]$

$\frac{1}{2}S = \sum_{k=2}^{\infty} \frac{ln(k) - ln(k-1)}{2^{k-1}}$

$\frac{1}{2}S = \sum_{k=2}^{\infty} \frac{ln(\frac{k}{k-1})}{2^{k-1}}$

$S = 2\sum_{k=2}^{\infty} \frac{ln(\frac{k}{k-1})}{2^{k-1}}$

Pull out the first term:

$S = 2 \cdot \frac{ ln\big(\frac{2}{1}\big)}{2} + 2\sum_{k=3}^{\infty} \frac{ln(\frac{k}{k-1})}{2^{k-1}}$

$S = ln(2) + 2\sum_{k=3}^{\infty} \frac{ln(\frac{k}{k-1})}{2^{k-1}}$

The numerator of the first remaining term in the series is $ln\big(\frac{3}{2}\big)$ . Because the series $\frac{k}{k-1}$ decreases monotonically toward $1$ for increasing values of $k$ , all of the remaining numerators in the series will be smaller than $ln\big(\frac{3}{2}\big)$ . By replacing every numerator in the remaining infinite sum with $ln\big(\frac{3}{2}\big)$ , the result will be something that is strictly greater than $S$ .

$S = ln(2) + 2\sum_{k=3}^{\infty} \frac{ln(\frac{3}{2})}{2^{k-1}}$

$S < ln(2) + 2 \cdot ln\Big(\frac{3}{2}\Big)\sum_{k=3}^{\infty} \frac{1}{2^{k-1}}$

Note that:

$\sum_{k=3}^{\infty} \frac{1}{2^{k-1}} = \frac{1}{2}$

Plugging this back in

$S < ln(2) + 2 \cdot ln\Big(\frac{3}{2}\Big) \cdot \frac{1}{2}$

$S < ln(2) + ln\Big(\frac{3}{2}\Big)$

$S < ln\Big(2 \cdot \frac{3}{2}\Big)$

$S < ln(3)$

$ln(P) < ln(3)$

$P < 3$

So the answer to the question is "no", there is no positive integer $n$ such that

$\sqrt{2\times\sqrt{3\times\sqrt{...\sqrt{(n-1)\times\sqrt{n}}}}}\hspace{3mm}>\hspace{2mm}3$