Determine the limit to which the following expression
Determine the limit to which the following expression tends as $n \to \infty$, for $\alpha > -1$: $$ a_n = \frac{1}{n^{1+\alpha}}\sum_{k=1}^{n} k^\alpha. $$
Explanation of Method
Rewriting the expression as $\frac1n\sum_{k=1}^{n}\left(\frac kn\right)^\alpha$ reveals it as a Riemann sum for $f(x)=x^\alpha$ on $[0,1]$ with $n$ equal subintervals. As $n\to\infty$, Riemann sums converge to the corresponding definite integral, so the limit is simply $\int_0^1 x^\alpha\,dx$ — which is finite precisely because $\alpha>-1$ keeps the integral from diverging near $x=0$.
- Rewrite $a_n$ as a Riemann sum for $f(x)=x^\alpha$ on $[0,1]$.
- Evaluate the corresponding definite integral.
- Verify the result against known closed forms for specific integer values of $\alpha$.
Step 1 · Recognize as a Riemann sum
$$ a_n = \frac{1}{n^{1+\alpha}}\sum_{k=1}^{n} k^\alpha = \frac1n\sum_{k=1}^{n}\left(\frac kn\right)^\alpha, $$
a Riemann sum for $f(x)=x^\alpha$ on $[0,1]$ with partition points $x_k=\frac kn$ and width $\Delta x=\frac1n$.
Step 2 · Convert to a Riemann integral
As $n\to\infty$:
$$ \lim_{n\to\infty} \frac1n\sum_{k=1}^{n}\left(\frac kn\right)^\alpha = \int_0^1 x^\alpha\,dx. $$
For $\alpha>-1$ (so that $\alpha+1>0$, keeping the integral finite at $x=0$):
$$ \int_0^1 x^\alpha\,dx = \left[\frac{x^{\alpha+1}}{\alpha+1}\right]_0^1 = \frac{1}{\alpha+1}. $$
Step 3 · Verify against known cases
$\alpha=0$: $a_n = \frac1n\sum_{k=1}^n 1 = 1$, matching $\frac1{0+1}=1$.
$\alpha=1$: $a_n = \frac1{n^2}\cdot\frac{n(n+1)}2 = \frac12+\frac1{2n}\to\frac12$, matching $\frac1{1+1}=\frac12$.
$\alpha=2$: $a_n = \frac1{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \frac13+\frac1{2n}+\frac1{6n^2}\to\frac13$, matching $\frac1{2+1}=\frac13$.
Summary Table
| $\alpha$ | $a_n$ (exact) | Limit |
|---|---|---|
| $0$ | $1$ | $1$ |
| $1$ | $\dfrac12+\dfrac1{2n}$ | $\dfrac12$ |
| $2$ | $\dfrac13+\dfrac1{2n}+\dfrac1{6n^2}$ | $\dfrac13$ |
| general $\alpha>-1$ | — | $\dfrac{1}{\alpha+1}$ |
Additional Notes
- This result is the discrete-to-continuous bridge underlying integral calculus itself: $\frac1n\sum_{k=1}^n f\left(\frac kn\right) \to \int_0^1 f(x)\,dx$ holds for any Riemann-integrable $f$, not just power functions — this problem is simply the special case $f(x)=x^\alpha$.
- The condition $\alpha>-1$ is essential: for $\alpha\le-1$, the integral $\int_0^1 x^\alpha\,dx$ diverges (e.g. $\alpha=-1$ gives $\int_0^1\frac1x\,dx=\infty$), so the limit would not be finite in that regime.
- Problem 19's series ($\frac1n\sum\ln k \to\infty$) can be seen as the "$\alpha\to0$ boundary case" of a related but distinct family — logarithms grow more slowly than any positive power, so its Riemann-sum analysis needs the extra $\ln n$ term subtracted out first, unlike the clean power-law case here.
Result
The limit of the expression is: