Suppose that $\lim_{n\to\infty}\frac{a_n}{b_n}$=1. And $\sum^{\infty}_{n=1}a_n$ converges, $\sum^{\infty}_{n=1}b_n$ diverges. Are such sequences $(a_n)$, $(b_n)$ exist?
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You may consider, for example, $a_n=\frac{(-1)^{n-1}}{\sqrt{n}}$ and $b_n=a_n+\frac{1}{n}$.
23rd
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If we assume that $\{a_k\}$ and $\{b_k\}$ are positive, no such sequences exist:
To see why, suppose that
$$\lim_{n \to \infty} \frac{a_n}{b_n}=1,$$
we may choose $N>0$ sufficiently large so that $b_n < 2a_n$ for all $n \geq N$. Then the tails of our series satisfy $$\sum_{k=N}^\infty b_k<\sum_{k=N}^\infty 2a_k,$$ so that $\sum b_k$ converges by the comparison test.
awwalker
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What if $b_n$ is not positive? – ՃՃՃ May 25 '13 at 05:39
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You assumed that they are positive sequences, and I did no say that they are positive. – Sandra West May 25 '13 at 05:40
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Yes, the proof works only for positive sequences. – Chung. J May 25 '13 at 05:46
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For $n \ge 3$, let $a_n= -\frac{1}{\log n}$ if $n$ is odd, and $\frac{1}{\log(n-1)}$ if $n$ is even. The series converges to $0$, consecutive terms cancel.
Let $b_n=-\frac{2}{\log 2n+\log n}$ if $n$ is odd, and $b_n=a_n$ if $n$ is even.
The sum of two consecutive terms, the first of which has odd index, of the series $\sum b_n$ is $\frac{1}{\log(n-1)}-\frac{2}{\log(2n)+\log(n)}$. This simplifies to $$\frac{\log(2n)+\log n-2\log(n-1)}{(\log(2n)+\log n)\log(n-1)}.$$
The top has limit $\log 2$. The bottom is slow-growing, so the series diverges.
Note that $\lim_{n\to\infty}\frac{a_n}{b_n}=1$.
Remark: The example by Landscape is much nicer.
André Nicolas
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