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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.

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The first inequality is not necessarily true, but what is true is that if the series on the right converges then so does the series on the left. That may be what you read.

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  • $\begingroup$ Thanks so much, I think that is what I read. $\endgroup$
    – Eric
    Commented Feb 12, 2013 at 0:39
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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.

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  • $\begingroup$ Thanks for the intuitive argument it really helps. $\endgroup$
    – Eric
    Commented Feb 12, 2013 at 1:15

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