What tools would you gladly recommend me for computing precisely the limit below? Maybe a starting point? limn→∞1log⁡(n)∑k=1ncos⁡(sin⁡(2πlog⁡(k)))k

Question

What tools would you gladly recommend me for computing precisely the limit below? Maybe a starting point?  limn→∞1log⁡(n)∑k=1ncos⁡(sin⁡(2πlog⁡(k)))k

trovabile4p · Accepted Answer

Damian Roberts · Accepted Answer

I would start with this:  0≤|1log⁡(n)∑k=0ncos⁡(sin⁡(2πlog⁡(k)))k|≤1log⁡n∑k=0n1k

RizerMix · Accepted Answer

Using the standard estimate for the harmonic numbers, |1log⁡n∑k=1ncos⁡(sin⁡(2πlog⁡(k)))k|≤1log⁡n∑k=1n1k =1log⁡n(log⁡(n)+γ+O(1/n)) So in the limit, |limn→∞1log⁡n∑k=1ncos⁡(sin⁡(2πlog⁡(k)))k|≤1 This is a pretty elementary bound, so you were probably already aware of it, but hopefully it helps.

What tools would you gladly recommend me for computing precisely the limit below? Maybe a...

Answered question

Answer & Explanation