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An interesting limit



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18 hours ago, BMAD said:

Hmmmm, my answer was the reciprocal of yours.  Maybe I am wrong.  Can you support your answer?

A quick check:


The second biggest term in the denominator is n^(n-1).  (n^(n-1))/ (n^n) = 1/n, so as n approaches infinity the second biggest term in the denominator is negligible compared to the biggest.  So are all the other terms in the denominator.  Since the biggest term in the denominator is the same as the one in the numerator (and since there are no negative terms in either), the limit won't end up being less than 1.

The whole #!:


As shown above, the only term that matters in the denominator is n^n.

Dividing all terms in the numerator by this makes it a sum of values that look like

((n-1)^n)/(n^n) = ((n-1)/n)^n.

Replacing the 1 with -a yields

((n+a)^n)/(n^n) = ((n+a)/n)^n = (1+a/n)^n.

Now, take the limit of this as n approaches infinity...

y = lim (1+a/n)^n

ln y = lim (n * ln(1+a/n))

ln y = lim (ln(1+a/n) / n^-1 )

As n approaches infinity, top and bottom approach 0... L'Hopitals!

ln y = lim   (1/ (1+a/n)) * (-a/n^2)   /   (-n^-2)

ln y = lim   (1/ (1+a/n)) * a

ln y = (1/(1+0)) * a

ln y = a

y = e^a

So ((n+a)^n) / n^n  approaches e^a in the limit.

Listing the terms in the numerator  in decreasing size is

n^n + (n-1)^n + (n-2)^n + ...

Dividing by the only term in the denominator that matters gives

1 + ((n-1)^n)/n^n + ((n-2)^n)/n^n + ...

Using the limit we found and substituting the values in for a gives

1 + e^-1 + e^-2 + ...

A geometric series with ratio 1/e.

The sum of this is then 1/(1-(1/e)) = 1/((e-1)/e) = e/(e-1).


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