For which x > 0 does the generalized ratio test imply convergence of \sum_{n=1}^\infty a_{n}?

ossidianaZ 2021-08-14 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

 \(\lim_{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\)

then \(\sum a_{n}\) converges,while if

\(\lim_{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\),

then \(\sum a_{n}\)  diverges. Let \(\displaystyle{a}_{{{n}}}={\frac{{{1}}}{{{1}+{x}}}}{\frac{{{2}}}{{{2}+{x}}}}\ldots{\frac{{{n}}}{{{n}+{x}}}}{\frac{{{1}}}{{{n}}}}={\frac{{{\left({n}-{1}\right)}!}}{{{\left({1}+{x}\right)}{\left({2}+{x}\right)}\ldots{\left({n}+{x}\right)}}}}\).

Show that \(a_{2 n}/a_{n} \leq e^{-x/2}/ 2\)  .

For which x > 0 does the generalized ratio test imply convergence of \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{{n}}}\)?

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Expert Answer

Derrick
Answered 2021-08-15 Author has 15485 answers
\(\displaystyle{\frac{{{a}_{{{2}{n}}}}}{{{a}_{{{n}}}}}}\leq{\frac{{{e}^{{-{\frac{{{x}}}{{{2}}}}}}}}{{{2}}}}\)
No, he ratio test does NOT imply the convergence for \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{{n}}}\).
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