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The $n$th term of a series is $$\ \frac{n}{\sqrt{n+1}}$$

Upon using the D' Almbert Ratio Test

I get $0$ as the limit implying convergence whereas the series seems divergent (each term greater than its respective term of the divergent series have $n$th term as $\sqrt n$)

Here's the working $\ \lim_{n\rightarrow \infty } \frac {u_{n+1}}{u_n}=\lim_{n\rightarrow \infty } \frac {(n+1)^{3/2}}{n(n+2)^{1/2}}=0$

Can somebody point out the error?

mathnoob123
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    The limit of the ratio is $1$, not $0$. The numerator and denominator both have order $n^\frac{3}{2}$. – quasi Jul 07 '17 at 10:09

2 Answers2

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$$\lim_{n \to \infty} \frac{(n+1)^\frac{3}{2}}{n(n+2)^\frac12}=\lim_{n \to \infty} \frac{n^\frac32(1+\frac1n)^\frac32}{n^\frac32(1+\frac2n)^\frac12}=\lim_{n \to \infty} \frac{(1+\frac1n)^\frac32}{(1+\frac2n)^\frac12}=1$$

Sahiba Arora
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$$\frac{u_{n+1}}{u_n} = \frac{\frac{n+1}{\sqrt{n+2}}}{\frac{n}{\sqrt{n+1}}} = \frac{(n+1)^\frac{3}{2}}{n\sqrt{n+2}} = \frac{n^\frac32(1+\frac{1}{n})^{\frac32}}{n^{\frac32}\sqrt{1+\frac2n}}$$

5xum
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