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How can I determine the value of the following limit?

$$\lim_{n\to\infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\left(1+\frac{4n}{n}\right)\right]^{1/n}$$

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

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    $\begingroup$ Take logarithms and view it in terms of a certain Riemann sum. $\endgroup$ – Zarrax Jun 3 '14 at 19:01
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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 $\frac{1}{n^3}$

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  • $\begingroup$ 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
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    $\begingroup$ 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

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