Interesting Limit to Infinity Problem

limn1n+2n+3n+...+nnnn\lim_{n\to\infty}\frac{1^{n}+2^{n}+3^{n}+...+n^{n}}{n^{n}}

alternatively,

limni=1n(in)n\lim_{n\to\infty}\sum_{i=1}^n{(\frac{i}{n})}^n

My friends and I have been having a lot of difficulty with this particular problem. We've tried a variety of methods, such as Riemann Integration and L'Hospital's rule, but we've been unsuccessful. After plugging in large values for n we have found an extremely clean answer that we are confident in, but we still have no proof? Any ideas?

Note by Logan Dymond
6 years, 2 months ago

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6 votes

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That's a very nice problem indeed! Here's my take.

Fix any positive integer k. Then for all n>k, the sum:

1n+2n++nnnn(nk)n+(nk+1)n++nnnn=(1kn)n+(1k1n)n++1. \frac{1^n + 2^n + \ldots + n^n}{n^n} \ge \frac{(n-k)^n + (n-k+1)^n + \ldots + n^n}{n^n} = \left(1 - \frac k n\right)^n + \left(1 - \frac{k-1}n\right)^n + \ldots + 1. (*)

From calculus, we know that for n>k, we have (1kn)nek \left(1 - \frac k n\right)^n \to e^{-k} as an increasing sequence in n. [ Proving that it's increasing is a bit tedious, e.g. you can replace n with a continuous variable x and differentiate with respect to x. ] So the RHS tends to ek+e(k1)++1e^{-k} + e^{-(k-1)} + \ldots + 1 as n tends to infinity. Taking the lim inf of (*) as nn\to\infty gives:

liminfn1n+2n++nnnn1+e1+e2++ek. \lim \inf_{n\to \infty} \frac{1^n + 2^n + \ldots + n^n}{n^n} \ge 1 + e^{-1} + e^{-2} + \ldots + e^{-k}.

Since this holds for each k, the LHS is at least 1+e1+e2+=ee11 + e^{-1 } +e^{-2} + \ldots = \frac e {e-1}.

On the other hand, for each n, we have:

1n+2n++nnnn=(1n1n)n+(1n2n)n++1e(n1)+e(n2)++1 \frac{1^n + 2^n + \ldots + n^n}{n^n} = \left(1 -\frac{n-1}n\right)^n + \left(1 - \frac{n-2}n\right)^n + \ldots + 1 \le e^{-(n-1)} + e^{-(n-2)} + \ldots + 1

Taking the limsup of both sides gives:

limsupn1n+2n++nnnnlimsupn(e(n1)+e(n2)++1)=ee1. \lim \sup_{n\to\infty} \frac{1^n + 2^n + \ldots + n^n}{n^n} \le \lim\sup_n (e^{-(n-1)} + e^{-(n-2)} + \ldots + 1) = \frac{e}{e-1}.

Comparing the lim inf & lim sup of both sides should give us ee1 \frac e {e-1}.

[ PS. Sorry for using lim inf and lim sup; I had wanted to apply Squeeze theorem directly, but somehow couldn't get it to work. ]

C Lim - 6 years, 2 months ago

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Hi sir... How did you come up with the inequality in the first statement?

Christian Baldo - 6 years, 2 months ago

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limn(in)n=limn(1nin)n=e(n1). \lim_{n\to \infty}(\frac{i}{n})^n =\lim_{n\to \infty} (1 -\frac{n-i}{n})^n = e^{-(n-1)}.

We get:

limni=1n(in)n= \lim_{n\to \infty} \sum_{i=1}^n {(\frac{i}{n})^n} = limni=1ne(ni)= \lim_{n\to \infty} \sum_{i=1}^n {e^{-(n-i)}} = limnee1ne1=ee1. \lim_{n\to \infty}{ \frac{e-e^{1-n}}{e-1}} = \frac{e}{e-1}.

---------------------------------------Q.E.D.---------------------------------------------

Is this not correct enough?

Pontus Hultkrantz - 6 years, 2 months ago

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Your first statement does not make any sense. The answer should be independent of nn, indicating that your answer is wrong.

You made the mistake of claiming that

limn(1f(n)n)n=ef(n) \lim_{n\rightarrow \infty} (1 - \frac{f(n) } { n} ) ^n = e^{- f(n) }

We know that the statement is true for constants, but not necessarily for functions. For example, if f(n)=n f(n) = n , then the expression is always 0, hence the limit is 0, and not en e^{-n} whatever that is. It might be true that

limn(1f(n)n)n=limnef(n), \lim_{n\rightarrow \infty} (1 - \frac{f(n) } { n} ) ^n = \lim_{n \rightarrow \infty} e^{- f(n) },

but that will require some work too (esp if limf(n) \lim f(n) does not exist).

Calvin Lin Staff - 6 years, 2 months ago

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Thanks, good point!

Pontus Hultkrantz - 6 years, 2 months ago

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Thematically it's quite similar to the solution provided by C L, but lacks sufficient mathematical rigor. Certainly your work vaguely suggests what the limit could be, as well as a means to prove it, but for the reasons Calvin has explained, it cannot stand as mathematically valid. One needs to be more careful--showing that the limit infimum is bounded below, in my opinion, is the critical step of C L's proof.

hero p. - 6 years, 2 months ago

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Thanks, good point!

Pontus Hultkrantz - 6 years, 2 months ago

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