Find limn→∞(n+1)n+1nn

Question

Find  limn→∞(n+1)n+1nn

immablondevl · Accepted Answer

Notice that for f(x)=xln⁡x you havef′(x)=ln⁡x2x−1xNow by mean value theoremf(n+1)−f(n)=f′(θn)for some θn such that n&amp;lt;θn&amp;lt;n+1 (by Mean Value Theorem). This means that for n→∞ we have θn→∞If we notice that limx→∞f′(x)=0 we get thatlimn→∞(n+1ln⁡(n+1)−nln⁡n)=limn→∞f′(θn)=0

Fallbasisz8 · Accepted Answer

Standard trick when things are difficult due to mixed terms like this: Write

y_{n} = \sqrt{n + 1} \ln (n + 1) - \sqrt{n} \ln (n + 1) + \sqrt{n} \ln (n + 1) - \sqrt{n} \ln n

Now the first two terms can be combined as

\ln (n + 1) (\sqrt{n + 1} - \sqrt{n}) = \ln (n + 1) \frac{1}{\sqrt{n + 1} + \sqrt{n}} \approx \frac{\ln n}{2 \sqrt{n}}

which tends to zero via any number of methods. The last two terms are handled similarly, from which you can deduce that the limit of

y_{n}

is zero, and the original limit is 1.

RizerMix · Accepted Answer

We can do a bit more. Starting with what you wrote ln⁡(yn)=ln⁡(n+1)n+1nn=n+1ln⁡(n+1)−nln⁡(n) rewrite n+1ln⁡(n+1)=n1+1n(ln⁡(n)+ln⁡(1+1n)) and now use Taylor series 1+1n=1+12n−18n2+O(1n3)ln⁡(1+1n) =1n−12n2+O(1n3) which makes n+1ln⁡(n+1)=n(1+12n−18n2+O(1n3))(ln⁡(n)+1n−12n2+O(1n3)) Now, replacing ln⁡(yn)=1n(1−12ln⁡(1n))+18(1n)3/2ln⁡(1n)+O(1n5/2) Now, using yn=eln⁡(yn) and Taylor series again yn=1+1n(1−12log⁡(1n))+O(1n) which shows the limit and also how it is approached.

Find \lim_{n\to\infty}\frac{(n+1)^{\sqrt{n+1}}}{n^\sqrt{n}}

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