BMAD 63 Report post Posted November 16 Find the Limit as n goes to infinity for: (1^n + 2^n + 3^n + 4^n....+ n^n) ---------------------over--------------------- (n^1+ n^2 + n^3 + n^4 ... + n^n) Share this post Link to post Share on other sites

0 EventHorizon 8 Report post Posted November 16 Spoiler e/(e-1) That is an interesting limit :-) Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted November 17 On 11/16/2018 at 6:48 AM, EventHorizon said: Hide contents e/(e-1) That is an interesting limit :-) Hmmmm, my answer was the reciprocal of yours. Maybe I am wrong. Can you support your answer? Share this post Link to post Share on other sites

0 EventHorizon 8 Report post Posted November 18 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: Spoiler 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 #!: Spoiler 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). Share this post Link to post Share on other sites

Find the Limit as n goes to infinity for:

(1^n + 2^n + 3^n + 4^n....+ n^n)

---------------------over---------------------

(n^1+ n^2 + n^3 + n^4 ... + n^n)

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