The following advanced exercise use a generalized ratio test to determine convergence of some series

alesterp 2021-05-05 Answered

The following advanced exercise use a generalized ratio test to determine convergence of some series that arise in particular applications, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if  $lim\left\{n\to \mathrm{\infty }\right\}\frac{a\left\{2n\right\}}{{a}_{n}}<\frac{1}{2}$ then $\sum {a}_{n}$converges,while if $lim\left\{n\to \mathrm{\infty }\right\}\frac{a\left\{2n+1\right\}}{{a}_{n}}>\frac{1}{2}$, then $\sum {a}_{n}$ diverges.

Let ${a}_{n}=\frac{1}{1+x}\frac{2}{2+x}\dots \frac{n}{n+x}\frac{1}{n}=\frac{\left(n-1\right)!}{\left(1+x\right)\left(2+x\right)\dots \left(n+x\right)}$.

Show that $\frac{{a}_{2n}}{{a}_{n}}\le \frac{{e}^{-x/2}}{2}$ . For which x > 0 does the generalized ratio test imply convergence of $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$?

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Expert Answer

Malena
Answered 2021-05-06 Author has 83 answers
$\frac{{a}_{2n}}{{a}_{n}}\le \frac{{e}^{-\frac{x}{2}}}{2}$
No, he ratio test does NOT imply the convergence for $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$.
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