1

Evaluate $$\lim_{n\to\infty}\left[\left(\frac1n\right)^n+\left(\frac2n\right)^n+\cdots+\left(\frac{n-1}n\right)^n+\left(\frac nn\right)^n\right]$$

I tried to solve this by taking logs on both sides and got this form: $$\log l = \lim_{n\to\infty} n\log\frac{n!}{n^n}=\lim_{n\to\infty}n\log\frac{n!}{n^n}$$ After applying limit, $\frac{n!}{n^n}\to0$, so $\log\frac{n!}{n^n}\to-\infty$ as $n\to\infty$. Then I am stuck. How to solve these type of problems?

Rócherz
  • 3,976
Mathaddict
  • 2,300

1 Answers1

6

Correct version of the question: $$\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^n=? $$ And this sum converges to $$ \lim_{n\to\infty}\sum_{k=0}^{\infty}\left(\frac{n-k}{n}\right)^n1_{k\leq n} = \sum_{k=0}^\infty \lim_{n\to\infty}\left(\frac{n-k}{n}\right)^n1_{k\leq n} = \sum_{k=0}^\infty e^{-k} = \frac{1}{1-e^{-1}}, $$ by (Lebesgue's) dominated convergence theorem or monotone convergence theorem. (Note: $(1-\frac{k}{n})^n \uparrow e^{-k}$ as $n\to \infty$.)

Myunghyun Song
  • 21,723
  • 2
  • 24
  • 60
  • I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n – Mathaddict Nov 18 '18 at 04:53
  • Sorry that I didn't explain the notation $1_{k\leq n}.$ This 'indicator' function takes value $1$ if $k\leq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0\leq k\leq n$. – Myunghyun Song Nov 18 '18 at 04:57
  • Okay, I got it. Thanks. – Mathaddict Nov 18 '18 at 05:04
  • 1
    There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously. – Myunghyun Song Nov 18 '18 at 09:30