Find the limit of (1+2+3+...n)^(1/n) as n goes to infinity

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In this video we use L'Hôpital's rule to calculate the limit of the sum of n numbers raised to 1/n as n goes to infinity!

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The proof seems pointlessly complicated. Given that we know that lim n->inf ( n^(1/n) ) = 1, which is a common limit, we can simplify the expression as such:
lim (1+2+3+...+n)^(1/n) =
= lim (n(n+1)/2)^(1/n) =
= lim =
= lim (n^(1/n)) * lim ((n+1)^(1/n)) * lim ((1/2)^(1/n)) =
= 1 * 1 * 1 = 1

If it is not clear that lim ((n+1)^(1/n)) = 1, then here is a short proof:
(n+1)^(1/n) >= (n)^(1/n) -> 1
(n+1)^(1/n) <= (2n)^(1/n) = (2)^(1/n) * (n)^(1/n) ->
-> 1 * 1 = 1

karolgajko

(1+2+3+...+n) = (n^2+n)/2 < n^2
lim n^2^(1/n) = lim n^(2/n) = 1
and use theorem about three series

Damjes

1:42 it would be nice to mention that you can take lim out of the ln because of contiuity and then apply L'Hospital

lovrovladic

Seems like a pointless calculation to me. n, n^2, n^700, it doesn't matter, n to any fixed power is not going to increase as fast as that 1/n in the exponent is going to bring it down to 1. It was obvious the limit would be 1 at first glance.

medexamtoolscom

Instead of applying L'Hopital's rule for the second time, can you not just divide the numerator and denominator by n and let n tend to infinity. You end up with a couple of 1/n terms which tend to zero, a couple of constants and an n in the denominator, meaning the whole thing tends to zero.

adamlea

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