If an=(1+2n)n , then find limn→∞(1−ann)n

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If an=(1+2n)n , then find  limn→∞(1−ann)n

Howard Gallagher · Accepted Answer

Due to  (1−ann)n=[(1−ann)n−an]−ahnn=[(1−ann)n−an]−an  limn→∞(1−ann)n−an=e  and limn→∞(−an)=e2  Let An=(1−ann)n−an and Bn=−an, by the claim below, you can get the result!

sineurosi0f · Accepted Answer

This is a very nice problem; but it does not file under &#039;&#039;composite limits&#039;&#039;, i.e., limits of the form x→ζf(g(x)).It is agreed that limn→∞an=limn→∞(1+2n)n=e2The function u↦log⁡(1−u) is analytic for |u|&amp;lt;1. Therefore &#039;&#039;by Taylor&#039;s Theorem&#039;&#039; there is a function u↦g(u), analytic for |u|&amp;lt;1, such thatlog⁡(1−u)=−u+u2g(u) (1)It follows that there is an M&amp;gt;0 such that|g(u)|≤MIf x and y are arbitrary real numbers with 0&amp;lt;|xy|&amp;lt;1 then we get from (1) thatlog⁡(1−xy)x=−y+xy2g(xy)Now put x=1n, y=an. Then the last equation saysnlog⁡(1−ann)=−ab+1nan2g(ann) (2)whereby this holds as soon as n is large enough to make |an|&amp;lt;nNow let n→∞. Then an→e2 and ann→0. Therefore the RHS of (2) converges to −e2. So does the LHS, and we get what we have conjectured all along:limn→∞(1−ann)n=exp⁡(−e2)

If a_n=(1+\frac{2}{n})^n , then find \lim_{n\to\infty}(1-\frac{a_n}{n})^n

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