Assume positive sequences $\{a_n\},\{b_n\}$ such that $a_n\leq a_{n+1}$ and $b_n\geq b_{n+1}$, $\lim_n b_n=0$ and $\sum_{n=1}^\infty a_n(b_n-b_{n+1})$ is convergent.
want to show $\lim_n a_n b_n=0$.
Any hint would be appreciated!
Product of an increasing sequence and decreasing sequence
3
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real-analysis
sequences-and-series
convergence
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0Question : That summation is finite implies it converges? Can we say $\sum_{n=1}^{\infty}\left(-1\right)^{n}$ is finite? – 2017-01-05
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0What do you mean by $<\infty$? Do you want to mean that the sum is finite? – 2017-01-05
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0There is the limit law $ \lim a_n b_n = (\lim a_n)(\lim b_n) $ – 2017-01-05
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0But one would have to prove the convergence of $\{a_n\}$ first @QthePlatypus – 2017-01-05
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0@CarlosIsraelJrl yes its a hint rather then the full solution – 2017-01-05
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0Question re-edited. Thanks for clarification. – 2017-01-05
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3@QthePlatypus I don't see your point. If I take $a_n=n$, $b_n=1/n^{100}$ then the limit law is of no help for this problem. – 2017-01-05
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0@ChenKe if you can show that $\lim a_n$ exists then you can go. $\lim a_n b_n = (\lim a_n)(\lim b_n) = (\lim a_n) 0 = 0 $ – 2017-01-05
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1Well the example I just gave you implies that you can never prove $a_n$ is even bounded... – 2017-01-05
1 Answers
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Since the sum converges, for all $\epsilon>0$, there is $N_0$ so that
$$\sum_{n=m}^\infty a_n (b_n - b_{n+1}) <\epsilon$$
for all $m\ge N_0$. Since $\{a_n\}$ is increasing,
$$\sum_{n=m}^\infty a_{m} (b_n - b_{n+1}) \le \sum_{n=N_0}^\infty a_n (b_n - b_{n+1}) <\epsilon.$$
But the first term is
$$\sum_{n=m}^\infty a_m (b_n - b_{n+1}) = a_{m}\sum_{n=m}^\infty (b_n - b_{n+1})=a_mb_m.$$
Thus $0\le a_mb_m <\epsilon$ for all $m\ge N_0$. Hence $\lim a_nb_n = 0$.