Evaluate $\lim_{n \to \infty} \frac{1^k+2^k+...+n^k}{n^k}-\frac{n}{k+1} $

Question

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.

Possible typos: the denominator is $n^{k+1}$ not $n^k$, and $y_n-y_{n-1}$ still has the same exponent $k+1$, not two different ones. — xbh, Oct 04 '18 at 09:55

LucaMac · Accepted Answer · 2018-10-04T10:12:09.193

3

$$ \lim\limits_{n \to +\infty} \frac{(k+1)\cdot(1+\ldots+n^k)-n^{k+1}}{(k+1)\cdot n^k} = \lim\limits_{n\to +\infty} \frac{(k+1)\cdot (n+1)^{k}+n^{k+1} - (n+1)^{k+1}}{(k+1)\cdot((n+1)^k - n^k)} = \lim\limits_{n \to +\infty} \frac{n^{k-1} \cdot (k(k+1))/2}{n^{k-1} \cdot k(k+1)} = \frac{1}{2}$$

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!

edited Oct 04 '18 at 10:12

answered Oct 04 '18 at 09:48

LucaMac

3,339

You are applying Stolz to $k$, while the limit is as $n\to\infty$. – Julián Aguirre Oct 04 '18 at 09:59
Not really, but it is still wrong! I'm trying to solve it! – LucaMac Oct 04 '18 at 10:06
Edited! Is it correct now? – LucaMac Oct 04 '18 at 10:13
how do you go from the LHS first line to RHS? Can you expand a little bit, please? Thank you – Naz Oct 04 '18 at 10:23
I am using Stolts Cesaro with $a_n = (k+1)(1+\ldots+n^k)-n^{k+1}$ and $b_n = (k+1)n^k$ since $b_n$ is increasing and $b_n \to +\infty$ – LucaMac Oct 04 '18 at 10:26
@LucaMac Yes, everything O.K. – Julián Aguirre Oct 04 '18 at 10:50
I need to explore my previous solution, as I was solving just like you (almost). In your first step I would also take out $(k+1)/(k+1)$ from fraction and then that would lead me to infinity. I must have been making a mistake somewhere. – Naz Oct 04 '18 at 12:02
How do you deal with $((n+1)^k-n^k)$ in denominator. I have: $(n^k-n^k + \mathcal{O}(n^{k-1}))$ – Naz Oct 04 '18 at 12:33
It's $n^k-n^k + kn^{k-1} + O(n^{k-2})$ – LucaMac Oct 04 '18 at 12:41
Would you mind writing out more steps between the second and the third steps, please? – Naz Oct 04 '18 at 13:05
Use Newton's expansion for $(n+1)^k = n^k + kn^{k-1} + O(n^{k-2})$ and $(n+1)^{k+1}=n^{k+1} + (k+1)n^k + k(k+1)/2 \cdot n^{k-1} + O(n^{k-2})$. Just put them in the limit and it should be clear – LucaMac Oct 04 '18 at 13:17

Sarthak Rout · Answer 2 · 2018-10-04T10:33:26.540

2

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 $\frac{1}{2}$ .

edited Oct 04 '18 at 10:33

answered Oct 04 '18 at 09:55

Sarthak Rout

367

It is easy to see that for $k=1$ the limit is $1/2$. The integral evaluates to $n/(k+1)+O(1)$. – Julián Aguirre Oct 04 '18 at 09:58
Integrals are not necessary to solve this problem. I would like to know an outline for a solution that does not involve them. Thank you – Naz Oct 04 '18 at 10:21
@JuliánAguirre You are right. Thanks – Sarthak Rout Oct 04 '18 at 10:32

score 2 · Answer 3 · answered Oct 04 '18 at 10:32

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$$ $$

Paramanand Singh · Answer 4 · 2018-10-04T11:12:25.630

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 $1/2$.

Evaluate $\lim_{n \to \infty} \frac{1^k+2^k+...+n^k}{n^k}-\frac{n}{k+1} $

4 Answers4

Linked