How can I determine the value of the following limit?


The first thing that came to my mind is approximation for $e$. But am not able to twist the expression accordingly. Please help.

  • 2
    $\begingroup$ Take logarithms and view it in terms of a certain Riemann sum. $\endgroup$ – Zarrax Jun 3 '14 at 19:01

Take the logarithm of the expression. You will get $\sum_{k=1}^{4n} \log (n+k) - \log n$ in the numerator. Can you handle from here?

EDIT OK so I'm getting $-3 \log n + \sum_{k=1}^{4n} \log (1+ \frac{k}{n}) \frac{1}{n} = -3 \log n + \int_{1}^{5} \log xdx$. The second term is a constant, and the second converges to $0$ at the rate $\frac{1}{n^3}$

  • $\begingroup$ Assuming k is the limit, $\log k=\frac{1}{n} [\log(n+1)+\log(n+2)+...+\log(n+4n)-4n\log n]$ $\endgroup$ – user141561 Jun 3 '14 at 18:58
  • $\begingroup$ you should use the fact that $\frac{\log n}{n} \to_n 0$ and $\sum_{k} \log (k+n) = n a$ where $a= \log 4$ if I'm not mistaken. Be careful with the algebra though. $\endgroup$ – Alex Jun 3 '14 at 19:01
  • $\begingroup$ I am really sorry. Can you type the whole solution please. $\endgroup$ – user141561 Jun 3 '14 at 19:07
  • 1
    $\begingroup$ Please see the edit $\endgroup$ – Alex Jun 3 '14 at 19:32
  • $\begingroup$ Where did you get $-3\log n$ from. I am confused! :( and i have 4 options. It was an MCQ question in exam. I'll post the options after 4-5 hours. $\endgroup$ – user141561 Jun 3 '14 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.