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I don't know how to simplify this expression to apply the ratio test to determine if the sequence converges or diverges

$$\lim_{n\to +\infty} \frac{\sqrt{n+1}}{2n^2+1}$$

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4 Answers

Hint: Use the fact that, if $\lim_{n\to \infty}\frac{a_{n+1}}{a_n} = b $ and $|b|<1$, then $\lim_{n\to \infty} a_n =0 $.

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I'd rewrite it as $$\cfrac{\sqrt{1+\frac1n}}{2\sqrt{n^3}+\sqrt{\frac1n}}.$$ You can actually show fairly directly that that converges, without need to resort to the ratio test.

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Can you please elaborate? –  Walker Mar 22 '13 at 23:12
    
Show that the numerator is bounded above by $2,$ the denominator bounded below by $2\sqrt{n^3},$ so the expression is bounded above by $\frac1{\sqrt{n^3}}.$ On the other hand, the expression is bounded below by $0.$ Apply Squeeze Theorem. –  Cameron Buie Mar 22 '13 at 23:31
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How about comparing with series $\displaystyle \frac{1}{n^{1.5}}$?? That should give you a finite answer when using the ratio test.

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Estimate sequence terms above and below $$\frac{1}{2(n+1)^{\frac{3}{2}}}=\frac{\sqrt{n+1}}{2(n+1)^2} \leqslant \frac{\sqrt{n+1}}{2n^2+1} \leqslant \frac{2\sqrt{n}}{2n^2}=\frac{\sqrt{n}}{n^2}=\frac{1}{n^{\frac{3}{2}}}.$$ Hence $$\frac{\sqrt{n+1}}{2(n+1)^2}=O({n^{-\frac{3}{2}}}), \;\;\;n\to\infty.$$

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