$\lim\limits_{n \to \infty }\sum_{i=1}^{n}{1 \over n+i-1} = \int_{1}^{2}{1 \over x}dx$
For the above equations, could one can prove it without using any derivatives?
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4divide by $n$ and switch indices to get $\frac{1}{n}\sum_{i=2}^n \frac{1}{1+\frac{i}{n}}$ which should look familiar – Brevan Ellefsen May 24 '17 at 02:42
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2@BrevanEllefsen Why not make that an answer? – Cameron Williams May 24 '17 at 02:44
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@CameronWilliams done :) – Brevan Ellefsen May 24 '17 at 02:45
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We can write this as a Riemann Sum by dividing by $n$ and switch indices to get $$\frac{1}{n}\sum_{i=1}^n \frac{1}{1+\frac{i-1}{n}}=\frac{1}{n}\sum_{i=2}^n \frac{1}{1+\frac{i}{n}}$$ Convert this to a Riemann Integral and use properties of integrals to get your form
Brevan Ellefsen
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would you elaborate the procedure regarding how to convert above summation into Riemann Integral form? – snapper May 24 '17 at 03:35
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it IS in riemann integral form. The limit change was to bring it into the form of a Riemann Sum, i.e. $\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^\infty f(k/n) = \int_0^1 f(x) dx$. In your case the bottom limit is $2$ and not $0$, but this is irrelevant here... to see this, pull the first two terms out of the sum and note that they go to zero because of the limit anyway. – Brevan Ellefsen May 24 '17 at 05:52