I have a question about this limit. Calculate$$\lim_{t\to{1-0}}(1-t)(\frac{t}{1+t}+\frac{t^2}{1+t^2}+...+\frac{t^n}{1+t^n}+...)$$ Can anyone help?
Evaluate the sum $\lim_{t\to{1-0}}(1-t)(\frac{t}{1+t}+\frac{t^2}{1+t^2}+...+\frac{t^n}{1+t^n}+...)$
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calculus
limits
1 Answers
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Let $h=-\log t$. Then
$$ \begin{eqnarray} \lim_{t\nearrow 1}(1-t)\sum_{k=1}^\infty\frac{t^k}{1+t^k} &=& \lim_{t\nearrow 1}\sum_{k=1}^\infty(1-t)\frac1{1+t^{-k}} \\ &=& \lim_{h\searrow 0}\sum_{k=1}^\infty(1-\mathrm e^{-h})\frac1{1+\mathrm e^{kh}} \\ &=& \lim_{h\searrow 0}\sum_{k=1}^\infty(h+O(h^2))\frac1{1+\mathrm e^{kh}} \\ &=& \int_0^\infty\frac1{1+\mathrm e^x}\mathrm dx \\ &=& \log2\;. \end{eqnarray} $$
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0Cool!How can you think of this? – 2012-04-20
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1@Gingerjin: I'll try to retrace my thinking. I noticed that $t^k/(1+t^k)=1/(1+t^{-k})$, so each term is just $1/(1+x)$ evaluated at different points $x_k=t^{-k}$, and those points move closer and closer together as $t\to1$, while the whole thing gets multiplied by something that goes to $0$. That suggested to view this as a sum that becomes an integral in the limit, but it wasn't quite right because the points weren't equidistant, so I worked out the transformation that made them equidistant. – 2012-04-20
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0@joriki I'm used to transforming infinite sums to integrals over finite intervals. How do you note the upper limit is necessarily $\infty$? How would you construct the $\Delta x_i$'s in such a case? – 2012-04-21
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1@Peter: The interval length here is $h$. I'm not sure I understand the question about $\infty$. The summation index $k$ runs up to $\infty$, so I don't see how the upper limit of the integral could be anything other than $\infty$. – 2012-04-21
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0@joriki Missed that. The thing is I usually use, let's say $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{1}{{n + k}}} = \int\limits_0^1 {\frac{{dx}}{{1 + x}} = \log 2} $$ and the upper limit depends on the partitions, $n$, not on the interval. – 2012-04-21
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0@Peter: I see, interesting. I guess those are different situations in that the summation limit occurs in the summand in one case but not the other. – 2012-04-21