How could I approach limn→∞(1+cos⁡(12n)2)n I tried several things to evaluate it, namely looking at it as cos⁡(12n+1)2n instead or as exp⁡(2n⋅ln⁡(cos⁡(12n+1)) and then trying to show that the limit of n⋅ln⁡(cos⁡(12n+1)) is 0 (for example using LHopitals rule), but I havent

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How could I approach  limn→∞(1+cos⁡(12n)2)n  I tried several things to evaluate it, namely looking at it as cos⁡(12n+1)2n instead or as exp⁡(2n⋅ln⁡(cos⁡(12n+1)) and then trying to show that the limit of n⋅ln⁡(cos⁡(12n+1)) is 0 (for example using LHopitals rule), but I havent

Jim Hunt · Accepted Answer

(1+cos⁡(12n)2)n≤(1+12)n=1  Now cos⁡u≥1−u22 for small u≥0, then  (1+cos⁡(12n)2)n≥(1−122(n+1))n→1

movingsupplyw1 · Accepted Answer

Continuing where you left off with LHopitals:  L=limn→∞nln⁡(cos⁡(12n+1))=limn→∞ln⁡(cos⁡(2−n−1))1n=limn→∞−tan⁡(2−n−1)⋅2−n−1ln⁡2−1n2  Supposing the limits are finite, we can break up the limit over multiplication like this:  ln⁡2limn→∞tan⁡(2−n−1)limn→∞n22n+1  The first limit is 0 by the continuity of tan⁡, and the second is easily shown to be 0 by repeatedly applying LHopitals.  So, our original limit is e2L=e0=1

Vasquez · Accepted Answer

You were very close to the right track. Write y=(1+cos⁡(ϵn)2)n⇒log⁡(y)=nlog⁡(1+cos⁡(ϵn)2) Now, using the series expansion of cos(t) for small angles (or equivalents cos⁡(ϵn)=1−12ϵ2n+… 1+cos⁡(ϵn)2=1−14ϵ2n+… log⁡(1+cos⁡(ϵn)2)=−14ϵ2n+… But ϵ=12 ; so log⁡(y)→0⇒y→1

How could I approach \lim_{n \to \infty} (\frac{1+\co

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