How can I write $\lim_{n\to\infty} \frac{1}{n^6}{(1+2^5+...+n^5)}$ as an Integral?

Question

How can I write this series as a definite integral? $\lim_{n\to\infty} \frac{1}{n^6}{(1+2^5+...+n^5)}$

We've just covered

$\lim_{n\to\infty}\int_{a}^{b}f_n=\int_{a}^{b}\lim_{n\to\infty}f_n$

When $(f_n)$ converges uniformly on (a,b) and each $f_n$ is Riemann inetgrable on (a,b)

And $\sum_{n=0}^{\infty}\int_{a}^{b}f_n=\int_{a}^{b}\sum_{n=0}^{\infty}f_n$

When $\sum(f_n)$ converges uniformly on (a,b) and each $f_n$ is Riemann inetgrable on (a,b)

My first guess was to write the limit as

$\lim_{n\to\infty} \sum_{k=1}^{n}\int_{0}^{1}\frac{k^5}{n^6}dx$

Which I believe can be re-written as

$\int_{0}^{1}(\lim_{n\to\infty} \sum_{k=1}^{n}\frac{k^5}{n^6})dx$

Then realised I was back to where I started. Can anybody give me some suggestions?

Answer 1

Hint: $\int_{0}^{1}x^5 dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \Big(\frac{k}{n} \Big)^5$

Answer 2

This is just the upper Riemann sum:

$$\sum_{i=1}^n \frac{1}{n}\left(\frac{i}{n}\right)^5$$

of $f(x)=x^5$ in $[0,1]$.

So let $f_n(x)=\left(\frac{\lceil nx\rceil}{n}\right)^5$. Then:

$$\int_0^1 f_n(x)\,dx = \frac1{n^6}\sum_{k=1}^n k^5.$$

Show that $f_n(x)\to x^5$.

Answer 3

This is much simpler: $\dfrac{1}{n^6}{(1+2^5+...+n^5)}=\dfrac1n\sum_{k=1}^n\Bigl(\dfrac kn\Bigr)^5\,$ is a Riemann sum.

Answer 4

$$ \frac{1}{n^6}\int_0^n x^5dx \le \frac{1}{n^6}{(1+2^5+...+n^5)} \le \frac{1}{n^6}\int_1^{n+1} x^5dx = \frac{1}{n^6}\int_0^{n} x^5dx+ \frac{1}{6n^6}(-1+(n+1)^6-n^6)$$

So

$$ \lim_{n\to\infty} \frac{1}{n^6}{(1+2^5+...+n^5)} = \lim_{n\to\infty} \frac{1}{n^6}\int_0^n x^5dx = \frac{1}{6} $$

Answer 5

So if you want to turn this into a Riemann sum, you're first going to want to decompose your sum into (something which tends to 0) * (a series which tends to definite values at its beginning and end), and that suggests: $$ \sum_{k=1}^n \frac 1 n \cdot \frac {k^5} {n^5},$$ where the "series part" goes from $0 \rightarrow 1$ and the "partition size" goes to 0. That is, we define $x_k = k/n$ and find $\Delta x_k = x_k - x_{k-1} = 1/n$ and we can therefore rewrite this as$$\lim_{n\rightarrow\infty} ~ \sum_{k=1}^n ~ x_k^5 ~ \Delta x_k = \int_0^1 x^5 ~ dx.$$