1
$\begingroup$

The question is to evaluate $$\lim_{n \to \infty} n^{\frac {-1}{2} (1+\frac{1}{n})(1^1\cdot 2^2\cdots n^n)^{\frac{1}{n^2}}}$$

I tried to take ln both sides and apply the formula for Riemann sum but I failed to bring it in the form required.Any hint to bring the limit into some known form shall be highly appreciated .Thanks.

  • 0
    navinstudent did u mean $\displaystyle \lim_{n \to \infty} n^{\frac {-1}{2} (1+\frac{1}{n})}(1^1\cdot 2^2\cdots n^n)^{\frac{1}{n^2}}$ – 2017-01-31

1 Answers 1

1

Since $$A_n=\left( 1+\frac{1}{n} \right)\left( \prod_{k=1}^n k^k\right)^{1/n^2} \ge 1$$ you have $$0 \le n^{-\frac{1}{2}A_n} \le n^{-\frac{1}{2}} = \frac{1}{\sqrt n} \to 0$$ so by the squeeze theorem your limit is $0$.

  • 0
    Thanks for your answer.Can you please explain how did you get $A_n \geq 1$ – 2017-01-31
  • 0
    It is a product of numbers larger than $1$. – 2017-01-31
  • 0
    Please, explain why you don't accept this solution. Aren't you confident with squeeze theorem? – 2017-01-31
  • 0
    Sorry for my previous comments.squeeze theorem is very well applicable here.Thanks. – 2017-01-31