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Is it true that if $(a_n)_{n=1}^\infty$ is any sequence of positive real numbers such that $$\lim_{n\to\infty}(a_n)=0$$ then, $$\sum_{n=1}^\infty \frac{a_n}{n}$$ converges?

If yes, how to prove it?

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3 Answers 3

up vote 6 down vote accepted

It is false. For $n\gt 1$, let $a_n=\dfrac{1}{\log n}$.

The divergence can be shown by noting that $\int_2^\infty \frac{dx}{x\log x}$ diverges. (An antiderivative is $\log\log x$.)

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No: take $a_n:=\frac 1{\log n}$, then $$\sum_{j=2}^{N-1}\frac 1{j\log j}\leqslant \int_2^N\frac 1{t\log t}dt=[\log(\log t)]_2^N\geqslant \frac{\log \log N}2$$

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The problem is equivalent to finding a positive increasing function $f(n)$ that approaches $+\infty$ slower than the harmonic series.

Given such a function, choose $a_n$ so that $\sum \frac{1}{na_n} = f(n)$. Then $\sum \frac{1}{na_n}$ diverges.

Or, for simpler formulas, take a continuous $f(x)$ and choose $a(x)$ so that $\frac{1}{xa(x)} = f'(x)$. Then $\int \frac{1}{xa(x)}$ diverges and consequently so does $\sum \frac{1}{na_n}$.

The most common choice is $f(x) = \log \log x$.

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