Evaluate $\displaystyle \lim_{n \to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$. What are the ways of counting such things? My last topic in school was Riemann integral, can I use it here?
Evaluate limit with sum
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limits
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2y$e$s, you $c$an use it here :-) – 2012-04-22
3 Answers
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You sure can!
$ \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{(k/n)} .$
The second version of writing the sum makes it clearer that it is the Riemann sum of $f(x) = 1/x $ obtained by dividing $[1,2]$ into $n$ pieces and setting up rectangles over those intervals. As such, your limit is $ \int^2_1 \frac{1}{x} dx = \log 2.$
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Yes, you can use it.
The technique is $\int_{k+1}^{k+2}\frac{1}{x}d x \leq \frac{1}{k+1}\leq \int_{k}^{k+1}\frac{1}{x}d x.$
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The sum equals (the sum of 1/k from k=1 to 2n) minus (the sum of 1/k from k=n+2 to 2n). Then take the limit.