\(\lim_{n\to\infty}\frac{1^{n}+2^{n}+3^{n}+...+n^{n}}{n^{n}}\)

alternatively,

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

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?

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## Comments

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

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

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

From calculus, we know that for n>k, we have \( \left(1 - \frac k n\right)^n \to e^{-k}\) as an

increasingsequence 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 \(e^{-k} + e^{-(k-1)} + \ldots + 1\) as n tends to infinity. Taking the lim inf of (*) as \(n\to\infty\) gives:\( \lim \inf_{n\to \infty} \frac{1^n + 2^n + \ldots + n^n}{n^n} \ge 1 + e^{-1} + e^{-2} + \ldots + e^{-k}.\)

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

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

\( \frac{1^n + 2^n + \ldots + n^n}{n^n} = \left(1 -\frac{n-1}n\right)^n + \left(1 - \frac{n-2}n\right)^n + \ldots + 1 \le e^{-(n-1)} + e^{-(n-2)} + \ldots + 1\)

Taking the limsup of both sides gives:

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

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

[

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.]Log in to reply

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

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Comment deleted Jun 12, 2013

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Applying the AM-GM inequality \[ \frac{x_1 + x_2 + \cdots + x_n}{n} \ge (x_1 x_2 \ldots x_n)^{1/n} \] with the choice \( x_i = (i/n)^n \) yields \[ \frac{1}{n} \sum_{i=1}^n \left(\frac{i}{n}\right)^n \ge \left( \prod_{i=1}^n \frac{i^n}{n^n} \right)^{1/n} = \prod_{i=1}^n \frac{i}{n} = \frac{n!}{n^n}. \] Multiplying both sides by \( n \) then gives \[ \sum_{i=1}^n \left( \frac{i}{n} \right)^n \ge \frac{n!}{n^{n-1}}, \] which tends to \( 0 \) as \( n \to \infty \). As this lower bound is looser than \( \frac{e}{e-1} \), there is no contradiction. The AM-GM inequality simply doesn't furnish a sufficiently tight bound in this case (and for those who are familiar with applying it, this fact should be evident without any calculation).

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When you applied AM-GM, the RHS shouldn't be n!, but \( n \frac {n!}{n^n} \).

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

We get:

\( \lim_{n\to \infty} \sum_{i=1}^n {(\frac{i}{n})^n} = \) \( \lim_{n\to \infty} \sum_{i=1}^n {e^{-(n-i)}} = \) \( \lim_{n\to \infty}{ \frac{e-e^{1-n}}{e-1}} = \frac{e}{e-1}. \)

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

Is this not correct enough?Log in to reply

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

You made the mistake of claiming that

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

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

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

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

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Thanks, good point!

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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.

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Thanks, good point!

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