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I calculated $ L= \lim_{n \to \infty} \frac {1^n + 2 ^n + \cdots + n^n }{n^n}$ in two different ways getting two different answers.

  1. $$L = \lim_{n \to \infty}\left[1 + (\frac {n-1}{n})^n + (\frac {n-2}{n})^n + \cdots + (\frac {n-n}{n})^n\right] = 1+ \frac{1}{e} + \frac {1}{e^2} + \cdots = \frac {e}{e-1}. $$
  2. $$L = \lim_{n \to \infty} n \times \sum_{r=1}^{n}\left(\frac {r}{n}\right)^n \times \frac {1} {n} = \lim_{n \to \infty} n \times \left[ \int_{0}^{1} x^n \ dx \right] = \lim_{n \to \infty} \frac{n}{n+1} =1. $$

Which one is wrong? Where am I going wrong?

Benjamin Wang
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Anirban
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    Both are wrong. You have to justify your steps. – geetha290krm Mar 19 '24 at 11:22
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    The first is wrong. The limit of the sum is the sum of the limit but it holds for sum of a finite order. In this case, the expression in "square bracket" has more and more summands and even though each of them tends to a limit the size of the error (may) add up to any value. – Salcio Mar 19 '24 at 11:23
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    Both are wrong although the answer is $1$ – MathStackexchangeIsNotSoBad Mar 19 '24 at 11:27
  • @MathStackexchangeIsNotSoBad but the answer is not 1. – Benjamin Wang Mar 19 '24 at 11:29
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    This is why it is a very bad habit to carry the limit operator in calculations. Always transform expressions and at the last moment examine its limit. Also never take a limit of something with $\cdots$ inside, use the sum notation. – zwim Mar 19 '24 at 11:31
  • @BenjaminWang why?? After taking $n^n$ common from the numerator, isn't it coming 1$?$ – MathStackexchangeIsNotSoBad Mar 19 '24 at 11:32
  • There are many questions and solution for this question. The 1st solution of $e/(e-1)$ is a "incomplete solution that leads to the correct numerical answer". – Benjamin Wang Mar 19 '24 at 11:33
  • By the way, I think the 2nd solution is wrong for subtle reasons: the mesh of the Riemann integral is supposed to tend separately to zero than the $1/n$ already provided. – Benjamin Wang Mar 19 '24 at 11:34
  • @Salcio - I agree with your point. But if I take n = 100, the sum is very close to $\frac{e}{e-1}$. I tend to think the second method is wrong - but can't figure where it goes wrong ( well I've jumped a few steps in writing it here). – Anirban Mar 19 '24 at 11:37
  • @BenjaminWang , can you explain your last statement in a bit more detail? – Anirban Mar 19 '24 at 11:40
  • @Anirban please check my answer. – Benjamin Wang Mar 19 '24 at 11:42
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    Multiduplicate: https://math.stackexchange.com/questions/553895/how-find-this-lim-n-to-infty-sum-i-1n-left-fracin-rightn?noredirect=1 and https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Csum_%7Br%3D1%7D%5E%7Bn%7D%5Cleft(%5Cfrac%20%7Br%7D%7Bn%7D%5Cright)%5En%24&p=1 – Anne Bauval Mar 19 '24 at 11:43

1 Answers1

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The first answer is wrong although numerically correct. Please check these posts.

The second answer is wrong because

$$ \begin{align} L&=\lim_{n\to\infty}n\sum_{r=1}^{n}\frac1n\left(\frac{r}{n}\right)^n\\ &\neq\lim_{n\to\infty}n \lim_{k\to\infty} \sum_{r=1}^{k}\frac1k\left(\frac{r}{k}\right)^n\\ &= \lim_{n \to \infty} n \int_{0}^{1} x^n \ dx \\ &= \lim_{n \to \infty} \frac{n}{n+1} \\ &=1. \end{align} $$

Benjamin Wang
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  • Why is $ lim_{n \to \infty} n \sum_{r=1}^{n} \frac{1}{n} (\frac{r}{n})^n \neq lim_{n \to \infty}[ n \int_{0}^{1} x^n dx ] ? $ – Anirban Mar 19 '24 at 11:58
  • @Anirban check the middle step. Do you notice that the $k$ has to be a different variable than $n$? – Benjamin Wang Mar 19 '24 at 12:13
  • Note that the first sequence of equalities also don't make sense: you exchange the limit and the sum, but the index of the sum is exactly the value diverging in the limit, so in going from line 1 to line 2, the LHS doesn't depend on $n$ while the RHS does. Also of course, the same issue with the final equality, where some nonconstant function of $n$ is equated to the constant $e/(e-1)$. – stochasticboy321 Mar 19 '24 at 12:17
  • @stochasticboy321 Exactly. That's why I put a question mark on the inequality. – Benjamin Wang Mar 19 '24 at 12:26
  • In that case you're mischaracterising the question. They just write that $ \lim \sum_{0 \le k \le n} (1-k/n)^n = \sum_{k = 0}^\infty e^{-k},$ nothing like the sequence of equalities you've written is asserted. Of course this needs justification, but I see no reason why you need to put words in the OP's mouth to make that point. – stochasticboy321 Mar 19 '24 at 12:29
  • @stochasticboy321 you make a good point. – Benjamin Wang Mar 19 '24 at 12:32
  • @BenjaminWang , from Riemann's theorem, the Riemann sum would converge irrespective of how the interval (0,1) is partitioned. So if I choose the infinitesimal interval length as $ \frac {1}{n} $, there shouldn't be a problem. IMO, the error comes after that. The final step should be: $ lim_{ n \to \infty, \delta \to 0}[n \times (\frac {n}{n+1} + \delta)] = 1+ lim_{n \to \infty} (n \times \delta) \neq 1). \ $ I am not good in latex typing - so kept the writing short in original problem as well as here – Anirban Mar 22 '24 at 07:21