Finding limits using definite integrals $\lim_{n\to\infty}\sum^n_{k=1}\frac{k^{4}}{n^{5}}$

Question

Find the limit of $\displaystyle\lim_{n\to\infty}\frac 1 {n^5}(1^4+2^4...+n^4)$ using definite integrals.

It's equal to: $\displaystyle\lim_{n\to\infty} \sum^n_{i=1}\frac 1 i$ but now I'm not sure how to turn it to an integral.

$\Delta x_i=\frac 1 n, f(x_i)=1$ so the integral would be: $\displaystyle\int 1dx$ ? How can I find the bounds?

Answer 1

Hint:

Summation of the series using definite integral:

$$\lim \limits_{n\to \infty }\frac{1}{n} \sum \limits^{h(n)}_{r=g(n)}f(\frac{r}{n})=\int \limits^{b}_{a}f(x)dx$$

Where

1.$$\sum \to \int$$

2.$$\frac{r}{n} \to x$$

3.$$\frac{1}{n} \to dx$$

4.$$a=\lim \limits_{n\to \infty }\frac{g(n)}{n}$$

5.$$b=\lim \limits_{n\to \infty }\frac{h(n)}{n}$$

Answer 2

Hint: rewrite the sum as: $$\frac1n\sum_{k=1}^n\Bigl(\frac kn\Bigr)^4.$$ This is an upper Riemann sum for the function $x^4$ on the interval $[0,1]$.

Answer 3

Draw the curve $y=x^4$. By area comparison, we have $$\int_0^n x^4\,dx\lt 1^4+2^4+\cdots+n^4\lt \int_0^{n+1} x^4\,dx.$$ It follows that $$\frac{1}{5}\lt \frac{1^4+2^4+\cdots +n^4}{n^5}\lt \frac{1}{5}\cdot \frac{(n+1)^5}{n^5}.$$ Since $\lim_{n\to\infty}\frac{(n+1)^5}{n^5}=1$, it follows by Squeezing that our limit is $\frac{1}{5}$.

Answer 4

That can be tackled also without integrals. Obviously: $$ 4!\binom{n}{4}\leq n^4 \leq 4!\binom{n+3}{4} \tag{1}$$ and by summing both sides over $n=1,2,\ldots,N$ through a well-known combinatorial identity we get: $$ 4!\binom{N+1}{5}\leq\sum_{n=1}^{N}n^4\leq 4!\binom{N+4}{5}\tag{2} $$ and by dividing both sides by $N^5$ and letting $N\to +\infty$: $$ \lim_{N\to +\infty}\frac{1}{N^5}\sum_{n=1}^{+\infty}n^4 = \frac{4!}{5!}=\color{red}{\frac{1}{5}}.\tag{3}$$

Answer 5

Although not asked by the OP, I thought it might be interesting to mention that the sum $\sum_{k=1}^n k^4$ is well known (http://mathworld.wolfram.com/PowerSum.html) and can be written

$$\sum_{k=1}^n k^4 =\frac{1}{5}n^5 +\frac12 n^4+\frac13 n^3-\frac{1}{30}n$$

Dividing by $n^5$ and letting $n\to \infty$ recovers the expected result

$$\lim_{n\to \infty}\frac{1}{n^5}\sum_{k=1}^n k^4 =\frac{1}{5}$$

Answer 6

More generally, and with no originality, let $f$ be a function with $f'(x) \ge 0$ and $f(x) > 0$ Then $f(n) \le \int_n^{n+1} f(x) dx \le f(n+1) $. (If $f'(x) \le 0$ and $f(x) > 0$, then the inequalities are reversed so $f(n) \ge \int_n^{n+1} f(x) dx \ge f(n+1) $.)

Summing for $n = 1$ to $N-1$, $\sum_{n=1}^{N-1}f(n) \le \int_1^{N} f(x) dx \le \sum_{n=1}^{N-1} f(n+1) $ or $ f(0) \le \sum_{n=1}^{N}f(n)-\int_1^{N} f(x) dx \le f(N) $. Therefore, if $\frac{f(N)}{\int_1^{N} f(x) dx} \to 0 $, since $\frac{\int_0^{1} f(x) dx}{\int_1^{N} f(x) dx} \to 0 $, $\frac1{\int_0^{N} f(x) dx}\sum_{n=1}^{N}f(n) \to 1 $.

Letting $f(x) =x^p $ with $p \ge 1$, $\frac{p+1}{N^{p+1}}\sum_{n=1}^{N}n^p \to 1 $ or $\frac1{N}\sum_{n=1}^{N}\left(\frac{n}{N}\right)^p \to \frac1{p+1} $.

Answer 7

Another solution that does not require integrals: using Cesaro-Stolz, $$ \lim_{n\to\infty}\frac{(1^4+2^4+\dots+n^4)}{n^5} = \lim_{n\to\infty}\frac{(n+1)^4}{(n+1)^5-n^5} = \cdots $$