0
$\begingroup$

I want to evaluate this limit :

$$\lim_{n\to \infty}\frac {\left((2n+1)(2n+2).....(2n+n)\right)^{1/n}}n$$

What I did was that I bought the $\frac 1n$ inside and divided throughout by $n^n$ which left me with the answer as $2$. But the correct answer is $\frac {27}{4e}$. How?

PS : forgive me I dont know MathJax or LaTex so :P

$\endgroup$
2
  • $\begingroup$ Note: I reformatted your post pretty heavily. Please check to make sure I didn't accidentally change your meaning. $\endgroup$
    – lulu
    Commented Oct 5, 2016 at 11:38
  • $\begingroup$ @lulu Thanks a lot! $\endgroup$
    – Ayush
    Commented Oct 5, 2016 at 11:56

1 Answer 1

0
$\begingroup$

I'm not going to tell you everything. But I'll guide you through the right steps. The mistake with your method is that the limit must be applied after multiplication not after Multiplication. so here you go. Take the n inside. Take log of the the product. Now you have a summation of log((2n+r)/n) from r=1 to n, and whole divided by n. As n tends to infinity , this is a integral expressed as sum of limits. Find the integral and solve it using appropriate limits (what could it be ?)

Have fun solving !!

PS. nit giving entire answer so as to not deprive you of the satisfaction of solving it. And sorry for my formatting, i dont know latex or mathjax either.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .