# summation inequality/limit for decreasing sequences

Let $$(w_n)_{n=1}^\infty$$ be a decreasing sequence of real numbers satisfying $$\lim_{n\to\infty}w_n=0\;\text{ and }\;\sum_{n=1}^\infty w_n=\infty.$$ Conjecture. For each $$M,N\in\mathbb{N}$$ we have $$\lim_{j\to\infty}\frac{\sum_{n=(M-1)j+1}^{(M-1)j+N}w_n}{\sum_{n=(M-1)j+1}^{Mj}w_n}=0.$$ Obviously this is true if $$M=1$$ and I suspect it is true for general $$M$$, but I'm stuck on proving it. The problem is that $$w_n/w_{n+1}$$ can be arbitrarily large. On the other hand, such large jumps must be spaced out to ensure that $$\sum w_n=\infty$$, and it seems like that might be enough to get the limit expression small. But a proof is eluding me, and maybe the conjecture isn't even true, my intuition notwithstanding.

Any help would be much appreciated. Thanks!

## 1 Answer

Let us take $$M = 2$$, $$N = 1$$.

Let us define $$w_n$$ by steps. Say we have already defined it up to $$m$$. Then let $$w_{m + 1} = w_m$$, $$k > \max(\frac{m}{w_m}, m)$$ and $$w_n = \frac{w_m}{m - 1}$$ for $$n = m + 2, \ldots, m + m$$. Then for $$j = m$$ we have $$\frac{\sum_{n=m+1}^{m+1} w_n}{\sum_{n=m+1}^{2m} w_n} = \frac{w_m}{w_m + \frac{w_m}{m - 1} \cdot (m - 1)} = \frac{1}{2}$$.

Also we have $$\sum_{n=m+1}^{m+m} w_m > 1$$, so our series diverge.