0

Let $$a_n=\frac{1^k+2^k+3^k+...+n^k}{n^{k+1}}$$

I need to find $$\lim _{n\to \infty }\left(a_n\right)$$

I know that $a_n=\frac{1^k+2^k+3^k+...+n^k}{n^{k+1}}=\frac{1}{n}\left(\frac{1^k}{n^k}+\frac{2^k}{n^k}+\frac{3^k}{n^k}+...+\frac{n^k}{n^k}\right)=\sum _{i=1}^n\:\frac{i^k}{n^k}\cdot \frac{1}{n}$

how can i continue from here ?

thanks

Jack D'Aurizio
  • 353,855
D.Z.O
  • 25
  • See https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 – lab bhattacharjee Dec 18 '17 at 08:04
  • $a_n$ is a Riemann sum for $\int_0^1 x^k,dx$. – Angina Seng Dec 18 '17 at 08:05
  • @LordSharktheUnknown so $\lim {n\to \infty }a_n=\lim :{n\to :\infty :}\frac{1^k+2^k+3^k+...+n^k}{n^{k+1}}=\lim :{n\to :\infty :}\frac{1}{n}\left(\frac{1^k}{n^k}+\frac{2^k}{n^k}+\frac{3^k}{n^k}+...+\frac{n^k}{n^k}\right)=\lim :{n\to :\infty :}\sum _{i=1}^n:\frac{i^k}{n^k}\cdot \frac{1}{n}=\int _0^1:x^kdx=\frac{1-0^{k+1}}{k+1}$ is that correct ? – D.Z.O Dec 18 '17 at 08:12
  • @D.Z.O Exactly! – Angina Seng Dec 18 '17 at 08:40

3 Answers3

1

$$a_n = \frac{1}{n^{k+1}}\sum_{j=1}^n j^k = \frac{1}{n}\sum_{j=1}^n \dfrac{j^k}{n^{k} }$$

So,

$$\lim_{n \to \infty}a_n =\int^1_0 j^k\ dj = \dfrac{1}{k+1}$$

user8277998
  • 2,666
0

The hint.

By Stolz we obtain: $$\lim_{n\rightarrow+\infty}\frac{1^k+2^k+3^k+...+n^k}{n^{k+1}}=\lim_{n\rightarrow+\infty}\frac{n^k}{n^{k+1}-(n-1)^{k+1}}=\lim_{n\rightarrow+\infty}\frac{n^k}{(k+1)n^k+...}=\frac{1}{k+1}.$$

Michael Rozenberg
  • 194,933
0

Another way:

$$(r+1)^{k+1}-r^{k+1}=\sum_{m=0}^k\binom{k+1}mr^m$$

If $\displaystyle S_u=\sum_{r=1}^nr^u$

$$\sum_{m=0}^k\binom{k+1}mS_m=\sum_{r=0}^n[(r+1)^{k+1}-r^{k+1}]=(n+1)^{k+1}$$

$\implies S_{k-1}=O(n^k)$

$$\implies\binom{k+1}kS_k=(n+1)^{k+1}$$

lab bhattacharjee
  • 274,582