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Suppose $f(1,i)>0$ is a strictly decreasing sequence of reals.
Let $f(k+1,n)=f(k,n+1)−f(k,n)$.

If $f(2m+1,n)$ is for all integers $m$, a strictly decreasing function in $n$ and $\lim_{n\rightarrow\infty}f(2m+1,n)=0$, must $\sum_{n>1}f(1,n)<\infty$?

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    Do you have an example where \sum_{n>1} f(1,n) converges?2011-12-01

2 Answers 2

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As a counterexample you could take $f(1,n)=1/n$ for $n\geq 1$. Using induction you can prove that $f(k,n)=\frac{(-1)^{k+1}}{n\cdot\ldots\cdot(n+k-1)}$ Then clearly $f(2 m+1,n)$ is strictly decreasing in $n$ for $m\in\mathbb{N}$ and $\lim_{n\rightarrow\infty}f(2m+1,n)=0$ but $\sum_{n>1}f(1,n)$ is divergent.

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Thoughts; the harmonic series diverges so consider f(k,n) = g(k)/n where g is a sequence that does not grow too quickly, etc such that the additive constraint holds.

That is I am guessing this is not true.

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    The problem is really findin$g$ suitable $g$ such that f(k+1,n) = f(k,n+1)-f(k,n) as everythin$g$ else is trivially met.2011-12-01