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Could you please give me some hint how to deal with this question:

Suppose $\left|\frac {a_{n+1}}{a_n}\right|\le c_n$ for each n and $c_n<1$.

May we conclude that $\left|\frac {a_{n+1}}{a_n}\right|\le c_n<1$ and by ratio test series $\sum a_n$ converges absolutely ? Or there is counter-example ?

Thanks.

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2 Answers 2

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Define a sequence $a_1 = 3$, and $a_{n+1} = a_n - \dfrac{a_n}{n}$ for $n \geq 1$ then $\dfrac{a_{n+1}}{a_n} = 1 -\dfrac{1}{n}$, and $\lim_{n \to \infty} |\dfrac{a_{n+1}}{a_n}| = 1$. This example shows that the above convergence test can not be used since the limit of the ratio could be $1$ while the ratio test requires that it be less than $1$ for the series to be convergent.

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  • $\begingroup$ And how could you define $c_n<1$ , please ? $\endgroup$
    – user97484
    Apr 20, 2014 at 8:54
  • $\begingroup$ Let $c_n=|\frac{a_{n+1}}{a_n}|$ $\endgroup$ Apr 20, 2014 at 8:57
  • $\begingroup$ And why $\sum a_n$ does not converge ? $\endgroup$
    – user97484
    Apr 20, 2014 at 9:15
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It doesn't matter that you have used the "intermediary variable" $c_n$. You have shown that $$|\frac{a_{n+1}}{a_n}| < 1$$ and if this is true also for $n \to \infty$ then, by the ratio test, the series converges absolutely.

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