# Sequences whose differences tend to $0$

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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$f(2,n) = f(1,n+1)-f(1,n)$ is negative? – GEdgar Dec 1 '11 at 14:47
I fixed this inconsitency – julenøtt Dec 1 '11 at 15:02
Do you have an example where $\sum_{n>1} f(1,n)$ converges? – Dimitrije Kostic Dec 1 '11 at 15:36

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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Yea I think 1/n is a counterexample – julenøtt Dec 1 '11 at 15:01
The problem is really finding suitable g such that f(k+1,n) = f(k,n+1)-f(k,n) as everything else is trivially met. – Adam Dec 1 '11 at 15:52