I'm here to ask you guys if my logic is correct. I have to calculate limit of this: $$\lim_{n\rightarrow\infty}\sqrt[n]{\sum_{k=1}^n (k^{999} + \frac{1}{\sqrt k})}$$ At first point. I see it's some limit of $$\lim_{n\rightarrow\infty}\sqrt[n]{1^{999} + \frac{1}{\sqrt 1} + 2^{999} + \frac{1}{\sqrt 2} \dots + n^{999} + \frac {1}{\sqrt n} }$$ And now is generally my doubt. Can i assume that if this limit goes to one then if limit of larger sequence goes to one too, then my original sequence goes to one too?
I came up with something like this.
if $$\lim_{n\rightarrow\infty} \sqrt[n]{\sum_{k=1}^n k^{1000}}$$ this limit goes to one ( cause obviously, expression under original sum is much lower than second sum ) then original limit goes to one too.
But i made it even simplier. $$\sum_{k=1}^n k^{1000} < n^{1001}$$, so... $$\lim_{n\rightarrow\infty} \sqrt[n]{n^{1001}}$$ this limit obviously goes to 1. Ans my final answer is that original limit goes to one too. I did some kind of bounding. But i'm not sure if i can do it. Any answers and tips are going to be greatly appreciated :-) Thanks.