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Suppose all $a_n$ are real numbers and $\lim_{n\to\infty} a_n$ exists.

What is the condition for the convergence( or divergence ) of the series $$ a_1 + a_1 a_2 + a_1 a_2 a_3 +\cdots $$

I can prove that $ \lim_{n\to\infty} |a_n| < 1 $ ( or > 1 ) guarantees absolute convergence ( or divergence ).
What if $ \lim_{n\to\infty} a_n = 1 \text{ and } a_n < 1 \text{ for all } n $ ?

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    $\begingroup$ You are doing a re-formulation of the ratio test. These numbers $a_n$ are the ratios of consecutive terms. $\endgroup$ – GEdgar Jun 28 '13 at 13:34
  • $\begingroup$ Yes, u r right. I forgot most of these ideas in calculus, so a simple question is asked -_-... $\endgroup$ – booksee Jun 28 '13 at 13:52
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What if $\lim_{n\to\infty}a_n=1$ and $a_n<1$ for all $n$?

Then the series may or may not converge. A necessary criterion for the convergence of the series is that the sequence of products

$$p_n = \prod_{k = 1}^n a_k$$

converges to $0$.

If the $a_n$ converge to $1$ fast enough, say $a_n = 1 - \frac{1}{2^n}$ ($\sum \lvert 1 - a_n\rvert < \infty$ is sufficient, if no $a_n = 0$), the product converges to a nonzero value, and hence the series diverges.

If the convergence of $a_n \to 1$ is slow enough ($a_n = 1 - \frac{1}{\sqrt{n+1}}$ is slow enough), the product converges to $0$ fast enough for the series to converge.

Let $a_n = 1 - u_n$, with $0 < u_n < 1$ and $u_n \to 0$. Without loss of generality, assume $u_n < \frac14$.

Then $\log p_n = \sum\limits_{k = 1}^n \log (1 - u_k)$. Since for $0 < x < \frac14$, we have $-\frac32 x < \log (1-x) < -x$, we find

$$ -\frac32 \sum_{k=1}^n u_k < \log p_n < -\sum_{k=1}^n u_k,$$

and thus can deduce that if $\sum u_k < \infty$, then $\lim\limits_{n\to\infty}p_n > 0$, so the series does not converge.

On the other hand, if $\exists C > 1$ with $\sum\limits_{k = 1}^n u_k \geqslant C\cdot \log n$, then $p_n < \exp (-C \cdot\log n) = \frac{1}{n^C}$, and the series converges.

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  • $\begingroup$ Thanks for your explanation. I find the complete answer to my question in wiki ratio test. $\endgroup$ – booksee Jun 28 '13 at 13:54
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I believe it's indeterminant. Consider $a_n=1-\frac{1}{n+1}$ and $a_n=1-\frac{1}{\log_2{n+2}}$. Both converge to 1, but the first one converges very quicky - and ends up resulting in the harmonic series, and the second one does so very slowly, and I believe it does converge.

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  • $\begingroup$ You are right. Thanks. $\endgroup$ – booksee Jun 28 '13 at 13:55
  • $\begingroup$ harmonic series is a neat example ! $\endgroup$ – booksee Jun 28 '13 at 14:03
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    $\begingroup$ the first $a_n$ should be $a_n = 1 - \frac{1}{n+1}$ $\endgroup$ – booksee Jun 29 '13 at 16:11

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