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$$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem

I tried this approach:
$$ \frac{1}{\sqrt{n^3+1}}\le\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+1}} $$ $$ \frac{1}{\sqrt{n^3+1}}<\frac{2}{\sqrt{n^3+2}}<\frac{n}{\sqrt{n^3+1}}$$ $$\vdots$$ $$\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+n}}<\frac{n}{\sqrt{n^3+1}}$$

Adding this inequalities:

$$\frac{n}{\sqrt{n^3+1}}\leq X_n<\frac{n^2}{\sqrt{n^3+1}}$$

And this doesn't help me much. How should i proced?

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4 Answers

up vote 3 down vote accepted

Since $$ X_n=\sum_{k=1}^n\frac{k}{\sqrt{n^3+k}}\ge\sum_{k=1}^n\frac{k}{\sqrt{n^3+n}}=\frac{n^2+n}{2\sqrt{n^3+n}}, $$ it follows that $$ \lim_{n \to \infty}X_n \ge \lim_{n \to \infty}\frac{n^2+n}{2\sqrt{n^3+n}}=\infty, $$ i.e. $\lim_{n \to \infty}X_n=\infty$.

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Hint:

Use the fact that: $$\sum_{i=1}^n{i}=\frac{n^2 +n}{2}$$ And: $$\frac{1}{\sqrt{n^3+1}}\le\frac{i}{\sqrt{n^3+i}}\le\frac{i}{\sqrt{n^3+1}}$$

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Hint: use $\frac{i}{\sqrt{n^3+n}} \le \frac{i}{\sqrt{n^3+i}} \le \frac{i}{\sqrt{n^3+1}}$. I even think the Squeeze theorem can be avoided.

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From the inequality $$n^3 \leqslant n^3+k \leqslant n^3+n \leqslant 2n^3, \;\; (1 \leqslant k \leqslant n)$$ we have $$ \dfrac{1}{\sqrt{2n^3}} \leqslant \dfrac{1}{\sqrt{n^3+k}} \leqslant \dfrac{1}{\sqrt{n^3}},$$ therefore $$\dfrac{n(n+1)}{2\sqrt{2n^3}} = \dfrac{1}{\sqrt{2n^3}} \sum\limits_{k=1}^{n}{k} \leqslant \sum\limits_{k=1}^{n} \dfrac{k}{\sqrt{n^3+k}} \leqslant \dfrac{1}{\sqrt{n^3}}\sum\limits_{k=1}^{n}{k}= \dfrac{n(n+1)}{2\sqrt{n^3}}. $$

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