4

I had to find the limit of the sequence

$$ a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}$$, where $ k $ is a natural number.

After applying the Stolz theorem, I was able to get here.

$$\lim \:_{n\to \:\infty \:}\left(\frac{\left(k+1\right)\left(n+1\right)^k-\left(\left(n+1\right)^{k+1}-n^{k+1}\right)}{\left(k+1\right)\left(\left(n+1\right)^k-n^k\right)}\right)$$

I tried different ways on continuing from here but unfortunately I am unable to get anywhere. I appreciate any kind of help.

Edit:

For users who may want a solution without the Big O notation, I was able to solve this limit by using the binomial theorem to calculate largest coefficient, since the numerator and denominator are polynomials in $n$, as a user below suggested.

Drilon Aliu
  • 91

3 Answers3

6

We have

$$a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=a_n =\frac{(k+1)(1^k+2^k+\cdot \cdot \cdot +n^k)-n^{k+1}}{(k+1)n^k}$$

and then by Stolz-Cesaro

$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k}$$

and since by binomial expansion

  • $(k+1)(n+1)^{k}=(k+1)n^k+k(k+1)n^{k-1}+O(n^{k-2})$
  • $(n+1)^{k+1}= n^{k+1}+(k+1)n^k+\frac{k(k+1)}2n^{k-1}+O(n^{k-2})$

we have

$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k} = \frac{k(k+1)n^{k-1}-\frac{k(k+1)}2n^{k-1}+O(n^{k-2})}{k(k+1)n^{k-1}+O(n^{k-2})}=$$

$$\frac{\frac12+O(1/n)}{1+O(1/n)}\to \frac12$$

Edit

As an alternative, directly by Faulhaber's formula since

$$1^k+2^k+\cdot \cdot \cdot +n^k = \frac{n^{k+1}}{k+1}+\frac12 n^k+O(n^{k-1})$$

we have

$$\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=\frac{n}{k+1}+\frac12+O(1/n) -\frac{n}{k+1}=\frac12+O(1/n) \to \frac12$$

user
  • 154,566
  • Thank you for your help! I currently am learning calculus and still haven't learned the big O notation, but I can't wait to do so and understand your solution. – Drilon Aliu Oct 10 '21 at 23:39
  • 1
    @AlexParker It is an useful notation to put togheter all the terms with order less or equal to $n^{k-2}$ which vanish after division by $n^{k-1}$. – user Oct 11 '21 at 09:03
3

The numerator and denominator are polynomials in $n$.

using the binomial theorem the largest coefficents can be calculated. They are

$$\frac{\frac{k(k+1)}{2}n^{k-1}+\cdots}{k(k+1)n^{k-1}+\cdots }\to \frac{1}{2}$$

Rene Schipperus
  • 39,526
  • Thanks! I kept using the binomial theorem, but there were a lot of terms and I gave up because I thought I wasn't getting anywhere. Now it is clear :) – Drilon Aliu Oct 10 '21 at 23:41
3

There is another way to show more than the limit itself if you are familiar with generalized harmonic numbers since $$\sum_{i=1}^n i^k=H_n^{(-k)}$$ and for large values of $n$ $$H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-k)$$ which makes $$a_n=\left(\frac{1}{2}+\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\frac{\zeta (-k) }{n^{k} }\sim \frac{1}{2}+\frac{k}{12 n}$$

Claude Leibovici
  • 260,315