Jtdylktuy

$ 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.

Hint: $$0leq n^5+(n+1)^5+dots+(2n)^5leq (n+1)cdot2n^5leq2ncdot (2n)^5=(2n)^6$$

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

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?