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!

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.