Proving $\lim_{n \rightarrow \infty} \frac{\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$

Question

How do we prove that $$\lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$$?

This type of terms appear in problems on limits, but I am unable to prove this.

Please help me out.

Related : http://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk1 and http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 — lab bhattacharjee, Sep 14 '14 at 04:23
See also http://math.stackexchange.com/questions/150391/evaluate-lim-limits-n-to-infty-frac-sum-k-1n-kmnm1 — Martin Sleziak, Sep 19 '14 at 08:54

score 8 · Accepted Answer · answered Sep 14 '14 at 04:18

8

As $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx,$$

we have $$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(\frac rn\right)^a=\int_0^1x^a\ dx$$

answered Sep 14 '14 at 04:18

lab bhattacharjee

274,582

What if the sum would be from $r=1$ to $n-1$ instead of $n$? – Novice Jul 15 '20 at 17:02
1

@Lover, $$-f\left(\dfrac nn\right)=?$$ – lab bhattacharjee Jul 15 '20 at 17:14
Yes, it is one. But still I don't see the right reasoning. My problem is that I have to show that $\lim_{n\to\infty} \dfrac{\sum\limits_{i=1}^{n-1} i^{a-1}}{n^a}=\dfrac{1}{a}$. The solution is very similar to this written above, but still I don't catch it because of this $n-1$ above the sum. I would appreciate any help. – Novice Jul 15 '20 at 17:23
1

@Lover, $$\sum_{r=1}^{n-1}f(r/n)=-f(n/n)+\sum_{r=1}^nf(r/n)$$ right? – lab bhattacharjee Jul 15 '20 at 17:25
Yes, I see. Now everything is clear, thanks a lot! – Novice Jul 15 '20 at 17:37
I would like to additionally ask if the result depends on value of $a$? – Novice Jul 15 '20 at 17:42
@Lover, What if $$a+1=0?$$ – lab bhattacharjee Jul 15 '20 at 17:43

score 2 · Answer 2 · answered Sep 14 '14 at 05:32

2

You can use L'hopital's rule as

$$ \lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}} = \lim_{n \rightarrow \infty} \dfrac{\displaystyle n^a}{(1+a)n^{a}} = \frac{1}{a+1}$$

answered Sep 14 '14 at 05:32

Mhenni Benghorbal

47,431

How have you evaluated the derivative of the numerator? – Novice Jul 15 '20 at 17:06

score 1 · Answer 3 · answered Sep 14 '14 at 04:35

1

It also follows from the Cesaro theorem.

answered Sep 14 '14 at 04:35

Rene Schipperus

39,526

score 1 · Answer 4 · edited Sep 15 '14 at 13:08

Look at the: $$\sum_{r=1}^nr^a=1^a+2^a+\dotsb+n^a$$ part. It turns out that this is a polynomial, whose leading term is $\dfrac1{a+1}r^{a+1}$. For example:

\begin{align} \sum_{r=1}^nr^1&=\frac12n^2+\frac12n\\ \sum_{r=1}^nr^2&=\frac13n^3+\frac12n^2+\frac16n\\ \sum_{r=1}^nr^3&=\frac14n^4+\frac12n^3+\frac14n^2\\ \sum_{r=1}^nr^4&=\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n\\ \sum_{r=1}^nr^5&=\frac16n^6+\frac12n^5+\frac5{12}n^4-\frac1{12}n^2 \end{align}

How to prove?

One thing to note is that if: $$\sum_{r=1}^nf(r)=S(r)$$ then: $$S(r)-S(r-1)=f(r)$$ (prove this!). Also note that $x^{a+1}-(x-1)^{a+1}$ is a polynomial whose first term is $(a+1)x^a$ (prove this too!). Can you use these facts to prove the statement in bold above? (Once you have this, it's easy to answer the stated problem.)

Proving $\lim_{n \rightarrow \infty} \frac{\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$

4 Answers4