# Closed form of $\sum\limits_{i=1}^n k^{1/i}$ or asymptotic equivalent when $n\to\infty$

Is there a "closed form" for $\displaystyle S_n=\sum_{i=1}^n k^{1/i}$ ? (I don't think so)

If not, can we find a function that is asymptotically equivalent to $S_n$ as $n\to\infty$ ?

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Cesaro (particular case of the Stolz-Cesaro theorem) says $$\lim_{n\rightarrow +\infty}a_n=a\quad\Rightarrow \quad \lim_{n\rightarrow+\infty}\frac{\sum_{i=1}^na_i}{n}=a .$$ I assume $k>0$ and $k\neq 1$ (as $S_n=n$ is trivial if $k=1$). Since $\lim_{n\rightarrow +\infty}k^\frac{1}{n}=1$, we get $$\lim_{n\rightarrow+\infty}\frac{\sum_{i=1}^nk^\frac{1}{i}}{n}=1\quad\Rightarrow\quad S_n=\sum_{i=1}^nk^\frac{1}{i}\;\;\sim\; n.$$ I doubt there is a closed form. But we can go further in the asymptotics. Recall that if $a_n\sim b_n$ and if $b_n$ is (eventually) positive with $\sum_{i=1}^{+\infty}b_i$ divergent, then $\sum_{i=1}^{n}a_i\sim \sum_{i=1}^nb_i$. That's again Stolz-Cesaro if you want. Now by first derivative of $k^x$ at $0$ $$k^\frac{1}{n}-1\sim\frac{\ln k}{n} \quad\Rightarrow\quad S_n-n=\sum_{i=1}^nk^\frac{1}{i}-1\sim\sum_{i=1}^n\frac{\ln k}{i}=\ln k\cdot H_n\sim\ln k\cdot \ln n$$ where $H_n$ is the $n$th harmonic number. Hence $$S_n=n+\ln k\cdot\ln n+o(\ln n).$$

Going up to the second derivative of $k^x$, we get $$k^\frac{1}{n}-1-\frac{\ln k}{n}\sim \frac{(\ln k)^2}{2n^2}.$$ So the resulting series converges and therefore, using $H_n=\ln n +O(1)$,

$$S_n-n-\ln k \cdot H_n=O(1)\quad\Rightarrow\quad S_n=n+\ln k\cdot \ln n+O(1).$$

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Nice work (+1). It'd be interesting to see a plot of $|S_n-n|$ vs $\log{n}$ for different values of $k$. – Ron Gordon Jun 25 '13 at 0:46
Note that with your second idea one can go as far as one wants in terms of the expansion. – Pedro Tamaroff Jun 25 '13 at 0:48
@RonGordon Thank you, Ron. That would take someone with better computing skills than mine... – 1015 Jun 25 '13 at 1:00
@PeterTamaroff I did my best with this simple idea, I think. – 1015 Jun 25 '13 at 1:13
Thanks! Where can I learn that stuff? (I'm a high school student) – user72870 Jun 25 '13 at 8:30

Just note that, for $k>1$

$$\displaystyle S(n,k)=\sum_{i=1}^n k^{1/i} \sim \int_{1}^{n}k^{1/x}dx$$

$$=-\ln(k)\Gamma(0, -\ln(k))-k+\ln(k)\Gamma\left(0, -\frac{\ln(k)}{n}\right)+nk^{1/n} \longrightarrow (1),$$

where $\Gamma(s,x)$ is the upper incomplete gamma function. For the other case $k<1$

$$\displaystyle S(n,k)=\sum_{i=1}^n k^{1/i} \sim \int_{0}^{n}k^{1/x}dx= n{k}^{1/{n}}+\ln \left( k \right) \Gamma \left( 0,-{\frac{\ln\left( k \right) }{n}} \right) \longrightarrow (2).$$

Here is a numerical example for $k=0.2$ and $n=2000$ the sum is

$$1988.361173.$$

and the approximation using $(2)$ is

$$1987.851634.$$

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$$\sum_{i=1}^n k^{\frac{1}{i}}=n+\ln(k)\ln(n)+\gamma\ln(k)+\sum_{r=2}^\infty\frac{\zeta(r)\ln(k)^r}{r!}+O(\frac{1}{n})$$

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Proof? :) I'd love to see where this one came from... – apnorton Jun 25 '13 at 12:22