# If two infinites series obey an inequality from $k = N,\ldots,\infty$ does this imply that the inequality is obeyed for $k=1,\ldots, \infty$?

If $$\frac{1}{k^{\log(k)}} < \frac{1}{k^2}$$ for $100 < k$, does this automatically imply that: $$\sum_{k=1}^\infty \frac{1}{k^{\log(k)}} < \sum_{k=1}^\infty \frac{1}{k^2},$$ or only for $$\sum_{k=100}^\infty \frac{1}{k^{log(k)}} < \sum_{k=100}^\infty \frac{1}{k^2}?$$ (Notice the change in the indices.) If the first inequality is true, (I think remember reading or hearing somewhere this is true but I cant find the source), can someone please give me an explanation of the intuition behind this? The second inequality is obviously true.

Edit: Just to clarify I am not asking for a proof or disproof of this fact, I am mainly concerned about the intuition behind it.

To answer your header question simply: no. If the inequality holds for all terms $k=n\to\infty$ then if the bigger one converges, so does the smaller. If the smaller diverges, so does the larger. If the inequality holds for ALL terms then the smaller convergent series converges to a value that is smaller than the larger series. If, however, the inequality is reversed for the first $n$ terms of the series, and both converge, then nothing can be said about the value of the "smaller" series - it could still converge to a larger value on account of the first terms being so much bigger.