Evaluate $\lim_{n \to \infty} \frac{1^k+2^k+...+n^k}{n^k}-\frac{n}{k+1} $ For some fixed natural $k$, we need to find the limit. I have been trying to solve this question in so many different ways now.. I will just outline one here. This should be solvable with the Stolz theorem.
My most recent idea was to treat the two terms separately, and first find the limit of the leftmost term. Then hopefully our limit is something that will take out the $n$ in the second term from the numerator. That is not the case...
Let's define $\tilde{x}_n = \sum_{i=1}^n i^k$, our numerator. Then $\tilde{y}_n = n^k$ will be our defnominator. We have that $\frac{x_n - x_{n-1}}{y_n - y_{n-1}} = \frac{n^k}{n^k - (n-1)^{k-1}}$, using binomial expansion or simply taking out the $n^k$, we see that this thing goes to plus infinity...
If I incorporate the second term in the original equation in my computations. Then I get that this difference ($\frac{x_n - x_{n-1}}{y_n - y_{n-1}}$) is also infinite: $$\frac{(1-n)(1-(1-1/n)^k)+k}{(k+1)(1-(1-1/n)^k}$$
So I am lost as to how to approach this problem now.
 A: $$ \lim\limits_{n \to +\infty} \frac{(k+1)\cdot(1+\ldots+n^k)-n^{k+1}}{(k+1)\cdot n^k} = \lim\limits_{n\to +\infty} \frac{(k+1)\cdot (n+1)^{k}+n^{k+1} - (n+1)^{k+1}}{(k+1)\cdot((n+1)^k - n^k)} = \lim\limits_{n \to +\infty} \frac{n^{k-1} \cdot (k(k+1))/2}{n^{k-1} \cdot k(k+1)} = \frac{1}{2}$$
The first equality is due to Stoltz Cesaro's Theorem, while in the second one I am just considering the highest power in $n$ since they are polynomials.
I hope now it is correct!
A: I am giving you a outline as to how approach the problem. The first term can be expressed as a definite integral by multiplying $n$ and dividing by it.  The integral evaluates to $\frac{n}{k+1}$. So you get the answer as $\frac{1}{2}$ .
A: Without Stoltz Cesaro's theorem, using generalized hrmonic numbers
$$\sum_{i=1}^n i^k=H_n^{(-k)}$$ Using the asymptotics
$$H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-k)$$
$$\frac{\sum_{i=1}^n i^k}{n^k}=\left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+n^{-k} \zeta (-k)$$
$$\frac{\sum_{i=1}^n i^k}{n^k}-\frac{n}{k+1}=\frac{1}{2}+\frac{k}{12
   n}+O\left(\frac{1}{n^3}\right)+n^{-k} \zeta (-k)$$
For illustration purposes, let us try using $k=5$ and $n=50$, the exact value should be $ \frac{762499}{1500000}\approx 0.5083326667$ while the expansion would give $\frac{40031249999}{78750000000}\approx  0.5083333333$$
$$
A: Consider the function $f(x) =x^k$ and use the following lemma

Lemma: If $f\in C^{1}[0,1] $ then $$\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx=\frac{f(1)-f(0)}{2}$$
  (proof available here)

The desired limit here is equal to $(f(1)-f(0))/2=1/2$.
If $k$ is a natural number then the proof is not that difficult and one can prove that $$\sum_{i=1}^{n}i^k=\frac{n^{k+1}}{k+1}+\frac{n^{k}}{2}+P(n)$$ where $P(n) $ is a polynomial in $n$ of degree less than $k$. The desired limit is then obviously $1/2$.
