Evaluate $\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$
Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$
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2This is Problem 2.3.12 (g) from Kaczor-Nowak: Problems in Mathematical Analysis I. The problem is stated on [p.37](http://books.google.com/books?id=HrO6QzUHU-gC&pg=PA37) and a solution is given on [p.184](http://books.google.com/books?id=HrO6QzUHU-gC&pg=PA184). This solution uses [Stolz-Cesaro Theorem](http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem). – 2012-05-25
6 Answers
Hint for another method:
You can use Faulhaber's formula: $1^p+2^p+\dots+n^p=\frac1{p+1}\sum_{k=0}^p\binom{p+1}kB_k\, n^{p+1-k},$ where $B_k$ denotes the $k$-th Bernoulli number (with the convention that $B_1=-\frac12$).
The result is more general.
Fact: For any function $f$ regular enough on $[0,1]$, introduce $ A_n=\sum_{k=1}^nf\left(\frac{k}n\right)\qquad B=\int_0^1f(x)\mathrm dx\qquad 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 says that $ 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 $f(1)-f(0)=C$.
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1+1. The all powerful Euler–Maclaurin!(http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula) – 2012-05-24
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 $p$.
Users Did and Ragib Zaman have provided excellent solutions. You might also want to look at Euler–Maclaurin formula which is of significance in this context.
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0@user17762: You need a transparent attitude, in mathematics. – 2013-12-09
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.
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0Thank you for that, Ragib! I changed my downvote to an upvote. – 2012-05-25
another method, using Stolz–Cesàro theorem: let ${ x }_{ n }=\left( p+1 \right) \left( { 1 }^{ p }+{ 2 }^{ p }+...+{ n }^{ p } \right) -{ n }^{ p+1 },{ y }_{ n }=\left( p+1 \right) { n }^{ p }$ $\lim _{ x\rightarrow \infty }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty }{ \frac { \left( p+1 \right) { \left( n+1 \right) }^{ p }-{ \left( n+1 \right) }^{ p+1 }+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { \left( n+1 \right) }^{ p }-{ n }^{ p } \right) } = } \\ =\lim _{ x\rightarrow \infty }{ \left( \frac { \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right) }{ 2 } { n }^{ p-2 }+...+1 \right) }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right) }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right) } \right) + } \\ +\frac { -{ n }^{ p+1 }-\left( p+1 \right) { n }^{ p }-\frac { p\left( p+1 \right) }{ 2 } { n }^{ p-1 }-...-1+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right) }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right) } $ let's cobmine all coefficients of n,then divide numerator and denominator by $n^{ p-1 }$ and define sum of the all terms no more -1 power with $o\left( \frac { 1 }{ n } \right) $ $\\ \\ \lim _{ x\rightarrow \infty }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty }{ \frac { \frac { p\left( p+1 \right) }{ 2 } +o\left( \frac { 1 }{ n } \right) }{ p\left( p+1 \right) +\left( \frac { 1 }{ n } \right) } =\frac { 1 }{ 2 } } \\ $