Suppose we have the positive real sequence $(a_n)_{n\in\mathbb{N}}$. Then is the implication $$\lim_{n\to\infty}(a_{n+1}+a_{n+2}+\dots+a_{2n})=0\implies\sum_{n=1}^\infty a_n \text{ is convergent}$$ true? I tried to prove that the sequence of partial sums is Cauchy because the statement is equivalent to $\forall\epsilon\gt0:\exists N:n\gt N\implies|S_{2n}-S_n|\lt\epsilon$ where $S_n=\sum_{k=1}^na_k$ but I couldn't get it to work. I feel that it may also be possible to apply the Stolz–Cesàro theorem here because the limit of $(S_{2n}-S_n)-(S_{2n-2}-S_{n-1})=a_{2n}+a_{2n-1}-a_n$ seems much simpler to study.
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Intuitively since Cauchy $\iff$ convergent you need $|S_m-S_n|<\epsilon$ for every $m$ but you only have it for $m=2n$. Then intuitively you can have $S_{2n}-S_n\sim \frac{1}{\log n}$ and simply have something like $S_{n^2} -S_n=S_{n^2}-S_{(n-1)n}+S_{(n-1)n}-S_{(n-2)n}+\cdots +S_{2n}-S_n\sim \frac{n}{\log n}$ so the sequence is not Cauchy so is not convergent we just need to find a sequence $S_{2n}-S_n \sim \frac{1}{\log n}$ we can indeed take $a_n=\frac{1}{n\log n}$ applying something like Stolz-Cesaro. This isn't rigorous that's why I'm leaving it as a comment; hope it helps. – kingW3 Oct 12 '19 at 16:29
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Let consider
$$a_n=\frac{1}{n \log n}$$
therefore
$$a_{n+1}+a_{n+2}+\dots+a_{2n} \le \frac{n}{(n+1) \log (n+1)}\to 0$$
but $\sum a_n$ diverges.
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1(+1) I had a feeling I was missing a trivial counter-example but I couldn't think of a series diverging slower than $\sum \frac1n$ – Peter Foreman Oct 12 '19 at 15:24
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@PeterForeman One way you could have thought of it is via the integral test: $\int\frac{dx}{x}=\ln|x|+C$ ($C$ locally constant), whereas $\int\frac{dx}{x\ln x}=\ln|\ln x|+C$. Obviously, a double logarithm diverges even more slowly. – J.G. Oct 12 '19 at 15:27
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@J.G. Yes I remember seeing this question now which provides arbitrarily "slow" diverging series. – Peter Foreman Oct 12 '19 at 15:35
