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Evaluation of $$\lim_{n\rightarrow \infty}\sqrt[n]{\sum^{n}_{k=1}\left(k^{999}+\frac{1}{\sqrt{k}}\right)}$$

$\bf{My\; Try::}$ First we will calculate $$\sum^{n}_{k=1}\left(k^{999}+\frac{1}{\sqrt{k}}\right)=n^{1000}\sum^{n}_{k=1}\left(\frac{k}{n}\right)^{999}\cdot \frac{1}{n}-\frac{1}{\sqrt{n}}\sum^{n}_{k=1}\sqrt{\frac{n}{k}}$$

Now How can I solve afeter that, Help me

Thanks in Advanced

juantheron
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2 Answers2

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Let $\displaystyle a_n=\sqrt[n]{\sum_{k=1}^{n}\Big(k^{999}+\frac{1}{\sqrt{k}}\Big)}$. Notice that $$1 <a_n<\sqrt[n]{2\sum_{k=1}^{n}k^{999}}=\sqrt[n]{2}~\sqrt[n]{\sum_{k=1}^{n}k^{999}}<\sqrt[n]{2}~\sqrt[n]{\sum_{k=1}^{n}n^{999}}=\sqrt[n]{2}~\sqrt[n]{n^{1000}}=\sqrt[n]{2}~(\sqrt[n]{n})^{1000},$$ and both sequences $\sqrt[n]{2},~(\sqrt[n]{n})^{1000}$ converge to $1$. Therefore, by the sandwich theorem, your limit exists and is equal to $1.$

Nikolaos Skout
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HINT

Note that for $k \ge 1$ we have $$ k^{999} + \frac{1}{\sqrt{k}} < 2k^{999} $$ and so your limit becomes $$ \lim_{n \to \infty} \sqrt[n]{2\sum_{k=1}^n k^{999}} = \lim_{n \to \infty} \sqrt[n]{2\Theta\left(n^{1000}\right)} = 1... $$

gt6989b
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