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How to use a definite integral to find the following limit: $\lim\limits_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right)$

Which is equal to $\ln(2)$. I know that there are a lot of series which converge with radius of convergence 1, so I'm assuming i'd have to use this fact in order to integrate the series, but I have no idea what to do here. If someone could drop a hint that would be great

David G. Stork
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  • The expression inside your parentheses is just a Riemann sum for the function $1/x$ on the interval $[n,2n]$ (or maybe $[n,2n+1]$ depending on how you count). Show that as $n$ gets large, this Riemann sum gets closer and closer to the actual integral on the associated interval. – Ethan Dlugie Apr 28 '20 at 22:29

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you can write the sum as :

$$ \sum_{n=k+1}^{2k}\frac1n=\sum_{n=k+1}^{2k}\frac1k\cdot\frac1{ \frac nk} $$

which is related to this integral via riemann sum :

$$\int_1^2\frac1x\,dx=\ln 2.$$

EDIT

It might be easier to see the Riemann sum in the following, equivalent expresion, as mentionned in the comments:

$$ \sum_{n=k+1}^{2k}\frac1n= \frac 1 n \sum_{n=1}^{k} \frac1 {{ 1+ \frac n k} } $$

which is the same expression as before, but with a substitution done.

Marine Galantin
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  • Not sure if your answer is really enlighten. I think that $$\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}$$ is more clear... which is (at the limit) $\int_0^1\frac{1}{1+x}dx$. Obviously, this will give the same result than you... – Todd Apr 28 '20 at 23:12
  • You're right, that's a good point. I won't edit my answer though, your comment is great :) – Marine Galantin Apr 28 '20 at 23:14
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    I downvoted. Not because your answer is wrong, but because is not so easy to read... honestly, reading your answer, I still don't understand your solution... I think I would rather call it a hard hint than a solution... – Todd Apr 28 '20 at 23:15
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    Strictly speaking, as you wrote it is not a Riemann sum. But indeed, doing a substitution, you can change it into a Riemann sum. – Todd Apr 28 '20 at 23:19