# Studies about $\sum_{k=1}^{n} x^{\frac 1k}$

Are there any studies about this function? $$f(x,n)=\sum_{k=1}^{n} x^{1/k}=x+x^{1/2}+x^{1/3}+x^{1/4}+\cdots +x^{1/n}$$

EDIT:

$f(1,n)=n$

$f'(1,n)=H_n$

$\int_0^1 \frac{f(x,n)}x\cdot dx=\frac{n^2+n}2$

where $f'(x,n)=\frac{d}{dx}f(x,n)$

To avoid fractional powers we can let $x=y^{n!}$ to change it into integer-powers, but the obtained polynomial does not have uniform paterm of powers.

$$f(y,n)=\sum_{k=1}^{n}y^{n!/k}=y^{n!}+y^{n!/2}+y^{n!/3}+y^{n!/4}+\cdots+y^{n!/n}$$ $$=y^{n!/n}\left(y^{n!-(n!/n)}+y^{(n!/2)-(n!/n)}+y^{(n!/3)-(n!/n)}+y^{(n!/4)-(n!/n)}+\cdots+y^{(n!/(n-1))-(n!/n)}+1\right)$$

As example let $n=4$ $$f(y,4)=y^6(y^{18}+y^6+y^2+y+1)$$

• I don't know about you, but using $a/b$ in exponents is more readable, since the exponents aren't very complicated here. Oct 18 '16 at 0:08
• You did the right edit my friend @SimpleArt Oct 18 '16 at 0:09
• I recommend you make it clear that $f'(1,n)$ refers to $\frac{d}{dx}f(x,n)$. It's not impossible to differentiate with respect to $n$ if one allows such extension. Oct 18 '16 at 0:09
• $f(x,n)\sim n$ as $n\to\infty$ Oct 18 '16 at 0:16
• $$\int_1^{n+1}x^{1/z}dz\le f(x,n)\le n+\int_1^nx^{1/z}dz$$for $x>1$ by the integral test. Oct 18 '16 at 0:27