# $\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$ using integral(lower-/upper- or Riemann-sums)

how can I show that $$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$$

using integral(lower-/upper- or Riemann-sums).

I tried the following:

$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}= \lim_{N\rightarrow\infty}\frac{1}{N^5} \lim_{N \rightarrow\infty} \sum_{n=1}^Nn^4=\frac{\frac{N^5}{5}}{N^5}=\frac{1}{5}$

But it feels like I am missing some steps here, especially why $\lim_{N \rightarrow \infty}\sum_{n=1}^N n^4=\frac{N^5}{5}$

How can I improve my try?

• Be careful as $\lim_{N \to +\infty} \sum_{n=1}^N n^4 = + \infty$. You probably mean: $$\sum_{n=1}^N n^4 \sim \frac{N^5}{5}$$ Commented May 1, 2018 at 11:45

Hint

$$\frac{\sum_{k=1}^n k^4}{n^5}=\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n} \right)^4=\frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n} \right)$$ with $f(x)=x^4$, so you can recognize a Riemann sum.

• Do you mean $f(x)=x^4$? Commented May 1, 2018 at 11:58
• Yes thank you :-). Commented May 1, 2018 at 11:59

It depends on what tools you have available.

Regardless, $$\lim_{n \to \infty}\dfrac{1^4+2^4+\cdots+n^4}{n^5}=\lim_{n \to \infty}\dfrac{\sum_{i=1}^{n}i^4}{n^5}\tag{*}$$

Method 1: You don't know of the Fundamental Theorem of Calculus.

I assume you have knowledge of the formulas \begin{align}\sum_{i=1}^{n}i &= \dfrac{n(n+1)}{2}\tag{1} \\ \sum_{i=1}^{n}i^2 &= \dfrac{n(n+1)(2n+1)}{6} \tag{2}\\ \sum_{i=1}^{n}i^3&=\left[\dfrac{n(n+1)}{2} \right]^2\tag{3}\end{align} Observe the formula $$\sum_{i=1}^{n}[(i+1)^5-i^5] = 2^5-1^5+3^5-2^5+\cdots+(n+1)^5-n^5=(n+1)^5-1\text{.}$$ From some FOILing, we find that $$(i+1)^5-i^5=5i^4+10i^3+10i^2+5i+1$$ so thus $$\sum_{i=1}^{n}[5i^4+10i^3+10i^2+5i+1]=5\sum_{i=1}^{n}i^4+10\sum_{i=1}^{n}i^3+10\sum_{i=1}^{n}i^2+5\sum_{i=1}^{n}i+\sum_{i=1}^{n}1$$ We have $\sum_{i=1}^{n}1 = n$, so inputting formulas $(1), (2), (3)$ into this equation, we obtain $$5\sum_{i=1}^{n}i^4+\dfrac{n(15n^3+50n^2+60n+31)}{6}=(n+1)^5-1$$ and solving for $\sum_{i=1}^{n}i^4$, we obtain $$\sum_{i=1}^{n}i^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$ Inputting this into $(*)$, we obtain $$\lim_{n \to \infty}\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=\lim_{n \to \infty}\dfrac{6n^5+15n^4+10n^3-n}{30n^5}=\dfrac{6}{30}=\dfrac{1}{5}\text{.}$$

Method 2: You do know the Fundamental Theorem of Calculus.

Recall that $\Delta x = \dfrac{b-a}{n}$, where $(a, b)$ is the interval of integration. Since there are no constants that stick out in equation $(*)$, let's arbitrarily set $b - a = 1$. From this, we obtain $\Delta x = \dfrac{1}{n}$, and the remaining terms correspond with $f(x_i)$: $$\lim_{n \to \infty}\dfrac{\sum_{i=1}^{n}i^4}{n^5}=\lim_{n \to \infty}\sum_{i=1}^{n}\dfrac{i^4}{n^5} =\lim_{n \to \infty}\sum_{i=1}^{n}\dfrac{i^4}{n^4}\cdot \dfrac{1}{n} = \lim_{n \to \infty}\sum_{i=1}^{n}\left(\dfrac{i}{n}\right)^4\Delta x$$ Thus, $f(x_i) = (i/n)^4$, so $x_i$ is probably $i/n$. The easiest way to obtain this formula for $x_i$ is to recall that $$x_i = a + i\Delta x = a + \dfrac{i}{n}$$ so it follows that $a = 0$ since $x_i = i/n$. Lastly, since $b - a = 1$, it follows that $b = a + 1 = 1$. Hence, we have $$\lim_{n \to \infty}\sum_{i=1}^{n}\left(\dfrac{i}{n}\right)^4\Delta x = \int_{0}^{1}x^4 \text{ d}x = \left.\dfrac{x^5}{5}\right|^{1}_{0}=\dfrac{1^5}{5}-\dfrac{0^5}{5}=\dfrac{1}{5}\text{.}$$

The expression could be rewritten as $$\lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^n \left(\frac{i}{n}\right)^4.$$ By the definition of Riemann integral, it should equal to $$\int_{0}^1 x^4 \,\mathrm{d}x = \frac{1}{5}.$$