Evaluate $ \lim \limits_{n \to \infty\ }\frac{1}{n}\sqrt[n]{n^5+(n+1)^5+...+(2n)^5}$

Question

$ \lim \limits_{n \to \infty\ }\frac{1}{n}\sqrt[n]{n^5+(n+1)^5+...+(2n)^5}$ I have no idea. Please give me some hint.

Answer 1

Hint: $$0\leq n^5+(n+1)^5+\dots+(2n)^5\leq (n+1)\cdot2n^5\leq2n\cdot (2n)^5=(2n)^6$$

Answer 2

1. Use $$ a^x = \exp( x \cdot \ln(a)) \ \forall x \in \mathbb{R} \ \forall a > 0. $$

2. Then use that $$ \lim_{n \to \infty} e^{f(n)} = \exp(\lim_{n \to \infty} f(n)), $$ since $x \mapsto e^x$ ist continuous on $\mathbb{R}$.

3. Lastly notice that the limit of a product is the product of the limits when both limits don't approach 0.

Alternative

Knowing (looking up...) $\sum_{k = 0}^{n} k^5 = \frac{n^2 (n + 1)^2 (2 n^2 + 2 n - 1)}{12}$ one may derive $$ \sum_{k = n}^{2n} k^5 = \frac{n^2 (n + 1) (42 n^3 + 24 n^2 + n - 1)}{4} \le \frac{n^2 \cdot n \cdot 42n^3}{4} \le 11 n^6, $$ where the latter inequalities will be true for large $n$, and proceed in a similar fashion from there

Answer 3

Hint :

$0 \lt (n^6)^{1/n} \lt((n+1)n^5)^{1/n} \lt $

$(n^5+(n+1)^5....+(2n)^{5})^{1/n} \lt$

$((n+1)(2n)^5)^{1/n}\lt ((2n)^6)^{1/n}.$

The lower and upper bound limits exist, hence

$(n^5+(n+1)^5+..(2n)^5)^{1/n} \lt M,$

where $M>0$, real.

The limit $n \rightarrow \infty $

$(1/n)(n^5+(n+1)^5+...(2n)^n)^{1/n}$

is?