# $\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluating this limit

Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$

-
 This is Problem 2.3.12 (g) from Kaczor-Nowak: Problems in Mathematical Analysis I. The problem is stated on p.37 and a solution is given on p.184. This solution uses Stolz-Cesaro Theorem. – Martin Sleziak May 25 '12 at 10:36

This is a nice little question. I am assuming that $p \in \mathbb{Z}^+$, though same could be said about it when $p \notin \mathbb{Z}^+$. Before getting to the answer lets experiment a bit for small positive integers $p$. To start off, you could try for some values $p$.

For $p=1$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n(n+1)}{2}}{n} - \frac{n}{1+1} \right) = \lim_{n \rightarrow \infty} \frac12 = \frac12$$

For $p=2$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n(n+1)(2n+1)}{6}}{n^2} - \frac{n}{2+1} \right) = \lim_{n \rightarrow \infty} \left(\frac{(n+1)(n+1/2)}{3n} - \frac{n}3 \right)\\ = \lim_{n \rightarrow \infty} \left(\frac{n}3 + \frac12 + \frac1{6n} - \frac{n}3 \right)= \frac12$$

For $p=3$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n^2(n+1)^2}{4}}{n^3} - \frac{n}{3+1} \right) = \lim_{n \rightarrow \infty} \left(\frac{n^2 + 2n + 1}{4n} - \frac{n}4 \right)\\ = \lim_{n \rightarrow \infty} \left(\frac{n}4 + \frac12 + \frac1{4n} - \frac{n}4 \right)= \frac12$$

Hence, we would guess that it is $\dfrac12$ independent of $p$. And this turns out to be right.

Let us denote $1^p + 2^p + \cdots n^p = P_p(n)$. This is a polynomial of degree $p+1$ and is given by $$P_p(n) = \frac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{p+1-k}$$ where $B_k$ are the Bernoulli numbers. These polynomials are related to the Bernoulli polynomials and there are some really nice results on these polynomials and more can be found here.

Hence, $$\dfrac{P_p(n)}{n^{p}} = \dfrac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{1-k} = \dfrac1{p+1} \left(B_0 n + (p+1) B_1 + \mathcal{O} \left(\frac1n\right) \right)$$ where $B_0 = 1$ and $B_1 = \frac12$. What you are looking for is $$\lim_{n \rightarrow \infty} \left(\dfrac{P_p(n)}{n^{p}} - \dfrac{n}{p+1} \right) = \lim_{n \rightarrow \infty} \left(\dfrac1{p+1} \left(n + (p+1) B_1 + \mathcal{O} \left(\frac1n\right) \right) - \dfrac{n}{p+1} \right)\\ = \lim_{n \rightarrow \infty} \left(B_1 + \mathcal{O} \left(\dfrac1n \right)\right)= B_1 = \frac12$$ independent of $p$.

Didier and Ragib Zaman have provided excellent solutions. You might also want to look at Eulerâ€“Maclaurin formula which is of significance in this context.

-
 I very much doubt the above very nice proof is appropriate for high school level. I don't think there are many education systems around the globe that include big "o" notation, Bernoulli numbers and polynomials in their H.S. curriculum... – DonAntonio May 24 '12 at 11:38 @DonAntonio The big O notation was only used as a short hand. It could have as well been written out. Also, Bernoulli numbers and polynomials are taught in high school. – user17762 May 24 '12 at 11:51 I agree about the big O notation. About Bernoulli numbers and pol's: I thoroughly know the education system in Israel and Mexico and, to a less extent, in USA. None of them includes anything even close to Bernoulli pol's/numbers in general (which means: perhaps some elite, most probably private high schools, do. I know none but I can't discard them). I also know a little about the system in Spain and it's the same as far as I know. – DonAntonio May 24 '12 at 11:56 @DonAntonio Ok. I did my schooling in India. And deriving $\sum_{k=1}^{n} k^p$ for $p = 1,2,3,4,5$ as a polynomial of degree $p+1$ was one of the popular exercises. We were not told that the coefficients are related to Bernoulli numbers but we were asked to find the pattern in the first few coefficients. – user17762 May 24 '12 at 12:00 Kudos for the indian h.s. education. In my globe's zone this kind of stuff is not even mentioned, let alone worked on. – DonAntonio May 24 '12 at 12:05

The result is more general.

Fact: For any function $f$ defined on $[0,1]$ and regular enough, introduce $$A_n=\sum_{k=1}^nf\left(\frac{k}n\right),\quad B=\int_0^1f(x)\mathrm dx,\quad C=f(1)-f(0).$$ Then, $$\lim\limits_{n\to\infty}A_n-nB=\frac12C.$$

For any real number $p\gt0$, if $f(x)=x^p$, one sees that $B=\frac1{p+1}$ and $C=1$, which is the result in the question.

To prove the fact stated above, start from Taylor formula: for every $0\leqslant x\leqslant 1/n$ and $1\leqslant k\leqslant n$, $$f(x+(k-1)/n)=f(k/n)-(1-x)f'(k/n)+u_{n,k}(x)/n,$$ where $u_{n,k}(x)\to0$ when $n\to\infty$, uniformly on $k$ and $x$, say $|u_{n,k}(x)|\leqslant v_n$ with $v_n\to0$. Integrating this on $[0,1/n]$ and summing from $k=1$ to $k=n$, one gets $$\int_0^1f(x)\mathrm dx=\frac1n\sum_{k=1}^nf\left(\frac{k}n\right)-\frac1n\int_0^{1/n}u\mathrm du\cdot\sum_{k=1}^nf'\left(\frac{k}n\right)+\frac1nu_n,$$ where $|u_n|\leqslant v_n$. Reordering, this is $$A_n=nB+\frac12\frac1n\sum_{k=1}^nf'\left(\frac{k}n\right)-u_n=nB+\frac12\int_0^1f'(x)\mathrm dx+r_n-u_n,$$ with $r_n\to0$, thanks to the Riemann integrability of the function $f'$ on $[0,1]$. The proof is complete since $r_n-u_n\to0$ and the last integral is $f(1)-f(0)=C$.

-
 +1. The all powerful Eulerâ€“Maclaurin!(en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula) – user17762 May 24 '12 at 11:47

If we draw the graph of $x^p$ from $x=1$ to $x=n,$ divide it into unit length intervals and approximate each segment of area by a trapezium (this is known as the trapezoidal rule) then we see that $$\int^n_1 x^p dx \approx \sum_{k=1}^n k^p - \frac{n^p+1}{2}.$$ The integral on the left is precisely $\displaystyle \frac{n^{p+1} -1}{p+1},$ so for large $n$ (where the major contribution is from the dominant terms) we have $$\sum_{k=1}^n k^p \approx \frac{n^{p+1}}{p+1} + \frac{n^p}{2}$$ so your limit is $1/2.$

For a precise solution, we need the error term along with the trapezoidal rule, which is derived here. It gives : $$\int^b_a f(x) dx = \frac{b-a}{2} ( f(a) + f(b) ) - \frac{(b-a)^3 }{12} f''(\zeta)$$ for some $\zeta \in [a,b].$ For $f(x)=x^p$ we have $f''(x) = p (p-1)x^{p-2}$ which is largest at $x=b$, the right end point. So the sum of the error terms in our application of the trapezoidal rule is at largest $$\frac{p(p-1)}{12} (2^{p-2} + 3^{p-2} + \cdots + n^{p-2}).$$ The sum in the brackets is overestimated by $\int^{n+1}_1 x^{p-2} dx= \frac{(n+1)^{p-1}-1}{p-1},$ so we get that $$\sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + E_n$$ where $E_n$ is an error term that satisfies $\displaystyle \lim_{n\to\infty} \frac{E_n}{n^p} = 0$ which proves your limit.

-
+1 Now this looks as something an advanced student in a very good (mathematicwise) high school could grasp. – DonAntonio May 24 '12 at 12:00
@Chris I just looked over my post and realized a horrible slip of the mind, it's no wonder why you didn't understand it before! The only wonder is how it got 7 upvotes in that condition lol. It's fine now though. Check out the edit and edit history for clarification. – Ragib Zaman May 25 '12 at 0:56
@RagibZaman Only the ideas are important. Yes you made a small algebraic error before but that doesn't really matter. :) – user17762 May 25 '12 at 3:50
I thought of $\int^n_1 x^p dx \approx \sum_{k=1}^n k^p - \frac{n^p+1}{2}$ but this doesn't modify the final result. – Chris's wise sister May 25 '12 at 10:17
@Ragib Zaman: i totally agree with what Marvis said: "Only the ideas are important". It's OK. – Chris's wise sister May 25 '12 at 10:21