An informal investigation of a sum.


Consider this sum:
$$S =\sum\_{k=2}^{\infty}(k^{1/k} -1)$$
Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.


Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S\_t = \sum\_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.


Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$,
$$S\_x=\sum\_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$


Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S\_2 = 4$ plus a remainder less than $1$. Going further $S\_3 = 9$ plus a remainder less than $1$, and $S\_{100} = 10,000$ plus a remainder less than $1$.
The output always seems to be the square plus a remainder! It seems true that $$\sum\_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x$ is the square root of any natural number greater than 1.
Here are some examples:
$S\_4 = 16.238932773$,
$S\_{12} = 144.5937831$,
$S\_{\sqrt{1729}} = 1729.84841$,
$S\_{50000} = 2500000000.988421705$.


Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.


Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$,
see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum\_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$
Let's define $$S\_t = \sum\_{k=2}^{\infty}\left(\sum\_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum\_{k=2}^{\infty}\sum\_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$
What is effectively adding column by column instead of row by row, we get:$$\sum\_{n=1}^{\infty}\sum\_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum\_{n=1}^{\infty}\frac{1}{n!}\sum\_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$


Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum\_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,


we get $$S\_t = \sum\_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$


Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.


$$\zeta(s) = \frac{1}{s-1} + \sum\_{n=0}^{\infty}(-1)^{n}\frac{\gamma\_n}{n!}(s-1)^{n}$$
Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma\_1 +\gamma\_2(s-1)-\frac12\gamma\_3(s-1)^2\cdots$$
$$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma\_2 -\gamma\_3(s-1)+\frac12\gamma\_4(s-1)^2\cdots$$
$$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma\_3 +\gamma\_4(s-1)-\frac12\gamma\_5(s-1)^2...$$
Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S\_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.


Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S\_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S\_x$.


$$\lim\_{x \to \infty}\Big|\Big(\sum\_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum\_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum\_{n=1}^{\infty}\frac{\gamma\_n}{n!} + \sum\_{k=3}^{\infty}\sum\_{n=k}^{\infty}(-1)^k\frac{\gamma\_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$


and $$\sum\_{n=1}^{\infty}\frac{\gamma\_n}{n!} = \frac12 - \gamma\_0,$$


and $$\lim\_{x \to \infty}\Big|\Big(\sum\_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum\_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum\_{k=3}^{\infty}\sum\_{n=k}^\infty(-1)^k\frac{\gamma\_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma\_0 = C$$ .


$\gamma\_n$ is bounded by $|\gamma\_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.


$$C = 0.988549601142268750644\ldots$$


So it seems $$\sum\_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C\_x,$$ where $-0.028501\ldots < C\_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C\_x$ when $x=1$. $C\_x$ is positive when roughly $x>~1.37$.


Edit: Hi everyone, I didn't know the strict format of this site. Please see the selected answer as a beautiful solution to my first question, and please consider the rest of this post as exposition of how the sum diverges.



The exploration of the asymptotic behavior of a simple sum. $\sum_{k=1}^{\infty} (k^{1/k} - 1)$
The OP asks the question whether the series $S =\sum\_{k=2}^{\infty}(k^{1/k} -1)$ diverges.


The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$
and $\sum\_{k=2}^\infty 1/k=\infty$.


