Suppose we have a sequence $\{\alpha_n\}$ such that $\alpha_n \ge 0$, $\sum_n \alpha_n = \infty$ and $\alpha_n \rightarrow 0$. Is it true that $\sum_n n^{-1} \, \alpha_n$ converges?
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1Take $a_n=\frac{1}{logn}$. – lulu Aug 18 '15 at 04:31
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To elaborate: $a_n=\frac {1}{logn}$ is a counterexample.
Most of the checks are trivial. To see that $\sum \frac {1}{nlogn}$ diverges compare it to $\int \frac{dx}{xlogx}$ to deduce that the partial sums go to $\infty$ like $log(log(x))$.
lulu
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We can prove more generally that if $\beta_n\ge0$ and $\sum_n\beta_n$ diverges (in your case $\beta_n=n^{-1}$), then there exists $\alpha_n$ with the given properties such that $\sum_n\beta_n\alpha_n$ diverges. To construct such $\alpha_n$, given $\beta_n$, construct a (strictly) increasing sequence $n_k$ such that $\sum_{j=n_{k-1}}^{n_k}\beta_j\gt1$. (This is possible since $\sum_n\beta_n$ diverges.) Then let $\alpha_n=k_n^{-1}$, where $k_n$ is the unique integer with $n_{k_n}\le n\le n_{k_n+1}$. Since $k_n\to\infty$ with $n\to\infty$, we have $\alpha_n\to0$, as required. But $\sum_n\alpha_n$ and $\sum_n\beta_n\alpha_n$ diverge, the first because $k_n^{-1}\ge n^{-1}$, the second because $\sum_{n=1}^{n_m}\beta_n\alpha_n\gt\sum_{k=1}^mk^{-1}$.
For more on this sort of argument and the lack of a boundary between convergent and divergent sequences, see this answer, and also this question.
