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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.

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  • $\begingroup$ Obviously Daniel, thanks for pointing that out... I mostly focused on good LaTeX representation and just made a stupid mistake. Besides of that, is it okay? $\endgroup$ – Krzysztof Lewko Dec 18 '13 at 15:22
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    $\begingroup$ There's one thing missing, you need to bound it from below (trivial, the sum is $\geqslant 1$, hence the $n$-th root is $\geqslant 1$). Without the lower bound, it could be that the limit is anything in $[0,1]$, so you need to state it. Then you have the squeezing $$1 \leqslant \sqrt[n]{\sum_{k=1}^n k^{999} + \frac{1}{\sqrt{k}}} \leqslant \sqrt[n]{2n^{1000}}.$$ $\endgroup$ – Daniel Fischer Dec 18 '13 at 15:27
  • $\begingroup$ Oh, yea it's true. Thanks for your help! $\endgroup$ – Krzysztof Lewko Dec 18 '13 at 15:36
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Because $$ \sum_{k=1}^nk^{999}\le\sum_{k=1}^n\left(k^{999}+\frac1{\sqrt{k}}\right)\le\sum_{k=1}^n\left(k^{999}+1\right) $$ and $$ \frac{n^{1000}}{1000}=\int_0^nx^{999}\,\mathrm{d}x\le\sum_{k=1}^nk^{999}\le\int_1^{n+1}x^{999}\,\mathrm{d}x=\frac{(n+1)^{1000}-1}{1000} $$ we have by the squeeze theorem, $$ \lim_{n\to\infty}\left(\sum_{k=1}^n\left(k^{999}+\frac1{\sqrt{k}}\right)\right)^{1/n}=1 $$

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    $\begingroup$ I see that you've corrected your answer to $1$. Your solution seems fine now. $\endgroup$ – robjohn Dec 18 '13 at 15:27
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Using the monotonicity of the function $x\mapsto x^\alpha$ we prove that:

$$\sum_{k=1}^n k^\alpha\sim_\infty\int_1^n x^\alpha dx\sim_\infty\frac{1}{\alpha+1}n^{\alpha+1}$$ hence we find $$\sqrt[n]{\sum_{k=1}^n (k^{999} + \frac{1}{\sqrt k})}\sim_\infty\left(\frac{n^{1000}}{1000}\right)^{1/n}\to1$$

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