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I'm new in Mathematical Analysis and now I have a problem in solving this problem:

Prove $$\lim_{n\to\infty}(\sin\frac{\ln2}{2}+\sin\frac{\ln3}{3}+\ldots+\sin\frac{\ln n}{n})^{\frac{1}{n}}=1$$

I think this problem can be solved by Squeeze Rule. But I have no idea how to construct the inequality relations it needs. Thanks soooo much for any hints!

Brain Zhang
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  • Please refrain from using displaystyle math in your question title. – AlexR Nov 25 '13 at 10:06
  • Just as an observation that might be helpful, the individual terms are always a number between ~0.5 and 0 and there are $n-1$ of them. – Dahn Nov 25 '13 at 10:14
  • When $\frac1n$ is in the exponent, I apply logarithm to use the formula mentioned here (http://math.stackexchange.com/questions/580393/evaluate-this-limit/580400#580400). But, exprssion is becoming prohibitively complex here – lab bhattacharjee Nov 25 '13 at 10:19

1 Answers1

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$\sin{\frac{\ln{2}}{2}}+\ldots+\sin{\frac{\ln{n}}{n}}\le1+\ldots+1=n-1$ and we know that $(n-1)^{\frac{1}{n}}\to1$ when $n\to\infty$. That's one inequality.

It should be that for every $n$ it is true that $0<\frac{\ln{n}}{n}<\frac{\pi}{2}$. So every term is actually positive and we can bound the sum with $\sin{\frac{\ln{2}}{2}}$ from below. And now just use that $constant^\frac{1}{n}\to1$ as $n\to\infty$.

Poppy
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