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Consider $\sum_{1}^{\infty} a_n$ , where $a_n$ =$\dfrac{(n+1)^n}{n^{n+\frac{3}{2}}}$

we find $$\dfrac{a_n}{a_{n+1}}=[\dfrac{(n+1)^2}{n(n+2)}]^n\times \dfrac{(n+1)^{\frac{5}{2}}}{(n+2)n^{\frac{3}{2}}}$$

by D-Alembert ratio test , $\lim_{n\rightarrow \infty}$ $\frac{a_n}{a_{n+1}}$ = $l$

i find $l= 1$, test fail,I have use Rabbes test, problem is very complicated, please help

Did
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user120386
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2 Answers2

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Note that, for every positive $n$, $$(n+1)^n=\left(1+\frac1n\right)^n\cdot n^n\leqslant\mathrm e\cdot n^n$$ hence $$a_n\leqslant\frac{\mathrm e}{n^{3/2}}$$

Did
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3

Note that

$$ a_n=\dfrac{(n+1)^n}{n^{n+\frac{3}{2}}} \sim \dfrac{(n)^n}{n^{n+\frac{3}{2}}}=\frac{1}{n^{3/2}}=b_n .$$

Now, we make comparison with the series $\sum_n b_n$ Check the limit

$$ \lim_{n\to \infty} \frac{a_n}{b_n}=e\neq 0 .$$

Then the two series converge together or diverge together.

Mhenni Benghorbal
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  • This takes for granted that $(n+1)^n\sim n^n$ (for an unspecified meaning of the symbol $\sim$), which is kind of the whole point. – Did Sep 02 '16 at 12:17