Evaluating $\lim_{k\to \infty} \frac{\sum_{n=0}^{k}x^{\frac{n}{k}}}{k+1}$ for all $x$ So I was really bored in my math class (I'm a high school sophomore taking precalculus in school) and my mind wandered to this expression:
$$\lim_{k\to \infty} \frac{\sum_{n=0}^{k}x^{\frac{n}{k}}}{k+1}$$
I have no idea how to even begin to evaluate this. I did some playing around with it on Desmos and it looks something like this for large values of $k$, which I don't recognize the shape of at all. I do notice that the graph should have the points $(0,0)$ and $(1,1)$, and it seems like if $j>k$ then $\frac{\sum_{n=0}^{k}x^{\frac{n}{k}}}{k+1}\ge\frac{\sum_{n=0}^{j}x^{\frac{n}{j}}}{j+1}$ for all $x$ with equality at $x=1$.
Also, please feel free to add any suitable tags---I didn't really know which ones to add.
(If you're wondering how I derived this expression, I was thinking about expressions like $\frac{x^0+x^1}{2}$, $\frac{x^0+x^{\frac{1}{2}}+x^1}{3}$, $\frac{x^0+x^{\frac{1}{3}}+x^{\frac{2}{3}}+x^1}{4}$, and then I thought about what would happen if we added more and more terms...)
 A: As already noticed by partial sum for geometric series for $x\neq 1$ we have
$$\sum\limits_{n=0}^k x^\frac{n}{k} =\sum\limits_{n=0}^k \left(x^\frac{1}{k}\right)^n = \frac{1 - x^{1+\frac1k}}{1 - x^{\frac1k}}$$
and therefore by standard limits
$$\frac{\sum\limits_{n=0}^k x^\frac{n}{k}}{k+1} = \frac{1 - x^{1+\frac1k}}{(k+1)\left(1 - x^{\frac1k}\right)}=\frac k{k+1}\frac{\frac1k}{x^{\frac1k}-1}\left( x^{1+\frac1k}-1\right)\to 1\cdot \frac1{\log x}\cdot (x-1)=\frac{x-1}{\log x}$$
A: For fixed $x \ne 1$, the quantity $\frac{1}{k+1} \sum_{n=0}^k x^{n/k}$ is an approximation of the integral $$\int_0^1 x^t \, dt = \frac{x-1}{\ln(x)} = \frac{1-x}{\ln(1/x)}.$$
By "approximation of an integral," I mean that if you consider the area under the function $f(t):=x^t$ from $t=0$ to $t=\frac{k+1}{k}$ and divide this interval into $k+1$ intervals each of width $\frac{1}{k}$, you can approximate the area under $f$ by summing the areas of $k$ rectangles each having area $\frac{1}{k} \cdot x^{n/k}$. See pictures of Riemann sums for more intuition. You are only in pre-calculus now, so you probably aren't expected to know this yet, but you will encounter this when you take calculus.
A: We can note $x^\frac{n}{k} = \left(\sqrt[k]{k}\right)^n$, and thus the sum in numerator is geometric series, and we have $$\sum\limits_{n=0}^k x^\frac{n}{k} = \frac{1 - \sqrt[k]{x}^{k + 1}}{1 - \sqrt[k]{x}} = \frac{1 - x \cdot \exp\left(\frac{\ln x}{k}\right)}{1 - \exp\left(\frac{\ln x}{k}\right)}$$
We have $\exp\left(\frac{\ln x}{k}\right) = 1 + \frac{\ln x}{k} + o\left(\frac{1}{k}\right)$ (I am not sure if you are expected to know this, and if not - what tools you can use).
After this substitution, we get under limit
$$\frac{\sum_{n=0}^{k}x^{\frac{n}{k}}}{k+1} = \frac{1 - x - \frac{\ln x}{k} + o(\frac{1}{k})}{-\left(\frac{\ln x}{k} + o\left(\frac{1}{k}\right)\right)\cdot (k + 1)} = -\frac{1 - x - \frac{\ln x}{k} + o(\frac{1}{k})}{\ln x + o(1)}$$
If $x \neq 1$, this simplifies to $-\frac{1 - x + o(1)}{\ln x + o(1)}$ with limit $\frac{x - 1}{\ln x}$.
If $x = 1$, we can find this sum directly - both numerator and denominator under limit equal $k + 1$, so limit is obviously $1$.
