<s>[INST] Limit with geometric sequence I computed
 $$\lim_{n \rightarrow \infty } \sqrt[n]{n^n+n^{n+1}+\cdots+n^{2n}} \cdot\left(1-\cos{\frac{3}{n}}\right)$$
$$=\lim_{n \rightarrow \infty } n^2 \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1} \cdot \left(1-\cos \frac{3}{n} \right) $$
$$=\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+ \cdots +1} \cdot \frac{1-\cos \frac{3}{n} }{\frac{9}{n^2}}\cdot 9=\frac{9}{2} $$
1) Is it correct to extract $n^{2n}$ from $\sqrt[n]{n^n+n^{n+1}+\cdots+n^{2n}}$ the way I did?
2)How do I show that $\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1}=1$?
I came up with this:
for geometric sequence the sum can be computed as $s_n=a_1\frac{q^n-1}{q-1}$, but I learned this in high school and if I recall correctly we used it for finite number of terms. So I don't know if I can use it here as $n\rightarrow \infty$. 
$$\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1} = \lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}\cdot \frac{n^n-1}{n-1}+1}$$
So I can estimate upper/lower bound:
$$\sqrt[n]{1} \le \sqrt[n]{n^{-n}\cdot \frac{n^n-1}{n-1}+1} \le \sqrt[n]n $$
Then:
$$\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1}=1$$
Thank you.
 [/INST] We have these inequalities
$$n^2\le(n^n+\cdots+n^{2n})^{1/n}\le((n+1)n^{2n})^{1/n}=n^2(n+1)^{1/n}\sim_\infty n^2$$
and  by Taylor series
$$1-\cos\left(\frac 3 n\right)\sim_\infty\frac 1 2\left(\frac{3}{ n}\right)^2=\frac{9}{2n^2}$$
hence
$$\lim_{n \rightarrow \infty } {\sqrt[n]{n^n+n^{n+1}+...+n^{2n}}\cdot(1-\cos{\frac{3}{n}})}=\lim_{n \rightarrow \infty }n^2\frac{9}{2n^2}=\frac{9}{2}$$
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