$$\lim _{n\to \infty \:}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}\right)$$
How I evaluate this limit?
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Related: http://math.stackexchange.com/questions/716/sum-of-the-alternating-harmonic-series-sum-k-1-infty-frac-1k1k – Mar 28 '15 at 19:56
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Hint: Consider writing this as a Riemann Sum. The limit in front then turns it into an integral of a function. – kvmu Mar 28 '15 at 19:56
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This limit is the same as $$ \lim_{x \uparrow 1} \lim_{n \to \infty} \left( x - \frac{x^2}{2} + \dotsb - \frac{x^{2n}}{2n} \right), $$ which can be written as $$ \lim_{x\uparrow 1} \lim_{n \to \infty} \int_0^x ( 1 - t + \dotsb -t^{2n-1} ) \, dt = \lim_{x\uparrow 1} \lim_{n\to\infty} \int_0^x \frac{1-t^{2n}}{1+t} \, dt $$ Since $x<1$, the $t^{2n}<x^{2n}$ tends uniformly to $0$ as $n \to \infty$, so we can interchange the limit and the integral to get $$ \lim_{x\uparrow 1} \int_0^x \frac{dt}{1+t} = \lim_{x\uparrow 1} \log{(1+x)}-0 = \log{2}. $$
Chappers
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Here: $ \lim_{x\uparrow 1} \lim_{n \to \infty} \int_0^x ( 1 - t + \dotsb -t^{2n} )$ , I think is $t^{2n-1}$ because we have $\frac{x^{2n}}{2n}$ – Andrei Mihai Mar 28 '15 at 20:51
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I thought that the limit was $$\lim_{n\rightarrow \infty}\lim_{x\rightarrow 1 -} \sum_{k=1}^{2n}\frac{(-x)^{k-1}}{k}$$ I think the real issue here lies in the fact that we actually can interchange the limits in your suggested way, but how does one motivate such a step? – TheOscillator Mar 28 '15 at 21:42
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Easiest way is probably Abel's theorem (http://en.wikipedia.org/wiki/Abel%27s_theorem) – Chappers Mar 28 '15 at 21:57