Find the limit as $n \rightarrow \infty$ of $\sum_{k=1}^{n} \frac{k^3}{n^4}$

Question

I would appreciate suggestions or hints for this homework question.

I'm asked to find the limit as $n \rightarrow \infty$ of the above series. The question is in a chapter on Riemann integration, so I assume that I need to formulate it as a Riemann sum (although I could be wrong about that).

So far, I have expanded the sum out to the partial, $S_{N} = \frac{1}{N^4} (1^3 + 2^3 + ... + (N-1)^3 + N^3)$.

But I'm not sure where to go from here.

Answer 1

a huge hint is that $1^3+2^3+...+N^3=\frac{(N(N+1))^2}{4}$

Answer 2

$$S_n=\frac 1n\Bigl((\frac 1n)^3+(\frac 2n)^3+...(\frac nn)^3\Bigr)$$

$$=\frac{1-0}{n}\sum_{k=1}^n(0+k\frac{1-0}{n})^3$$ $$=\frac{1-0}{n}\sum_{k=1}^nf(0+k\frac{1-0}{n})$$

with $ f(x)=x^3$.

$ f $ is continuous and integrable at $ [0,1] $, so $$\lim_{n\to+\infty}S_n=\int_0^1x^3dx$$ $$=\Bigl[\frac{x^4}{4}\Bigr]_0^1=\frac 14$$.

Answer 3

Use Stolz-Cesaro (look it up).

In this case it says $\lim_{n \to \infty} \dfrac{\sum_{k=1}^n k^3}{n^4} =\lim_{n \to \infty} \dfrac{n^3}{n^4-(n-1)^4} =\lim_{n \to \infty} \dfrac{n^3}{4n^3-6n^2+4n-1} =\dfrac14 $.

Answer 4

Making the problem more general $$S_p=\sum_{k=1}^{n} \frac{k^{p-1}}{n^p}$$ Faulhaber's formula write $$\sum_{k=1}^{n} k^{p-1}=\frac {n^p} p+\frac {n^{p-1}}2+\cdots$$ that is to say that $$S_p \sim \frac{\frac {n^p} p+\frac {n^{p-1}}2+\cdots }{n^p}=\frac 1 p+\frac 1 {2n}+\cdots$$ and so, you have not only the limit but also how it is approached when $n\to\infty$.