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 Calculus Calculus Math Forum

 November 28th, 2014, 01:23 AM #1 Member   Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1 Why doesn't this work? The task is to find this sum: And first I try to find an expression for f(x): Using partial fractions, I deduced (Edit: I see why this is wrong) Using Wolfram alpha, I see that the answer contains logarithm. The task also hints that you should use the fact that Integral[ln x dx] = xlnx - x + C. But I can't see how you can possibly mix an integral into this. Thanks in advance! November 28th, 2014, 08:21 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus $$\sum_{n=2}^{\infty}\frac{3}{n(n-1)5^n}=3\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)\cdot \frac{1}{5^n} =$$ $$=3\left(\sum_{n=2}^{\infty}\frac{\left(\frac{1}{ 5}\right)^{n}}{n-1}-\sum_{n=2}^{\infty}\frac{\left(\frac{1}{5}\right)^ {n}}{n}\right)=3\left(\frac{1}{5}\sum_{n=1}^{ \infty}\frac{\left(\frac{1}{5}\right)^{n}}{n}-\sum_{n=1}^{ \infty}\frac{\left(\frac{1}{5}\right)^{n}}{n}+ \frac{1}{5}\right)=$$ $$=3\left(\frac{1}{5}-\frac{4}{5}\sum_{n=1}^{\infty}\frac{\left(\frac{1} {5}\right)^{n}}{n}\right)$$ Remember that $$\log(1-x)=\int\frac{-1}{1-x}\;\mathbb{d}x=-\int\sum_{k=0}^{\infty}x^k\;\mathbb{d}x = -\sum_{k=1}^{\infty}\frac{x^{k}}{k}$$ for $$-1\leq x <1$$. So for $$x= \frac{1}{5}$$ we get, $$\sum_{n=2}^{\infty}\frac{3}{n(n-1)5^n}=\frac{3}{5}\left(1-4\sum_{n=1}^{\infty}\frac{\left(\frac{1}{5}\right) ^{n}}{n}\right)=\frac{3}{5}\left(1+4\log\left( \frac{4}{5}\right)\right)$$. Thanks from greg1313, topsquark, fysmat and 2 others Last edited by ZardoZ; November 28th, 2014 at 08:26 AM. November 28th, 2014, 02:15 PM #3 Member   Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1 Great, thanks! Thanks from ZardoZ November 28th, 2014, 07:00 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus I double checked with W|A and it gives as a result $$\frac{3}{5}\left(1-4\log\left( \frac{5}{4}\right)\right)$$. Of course this is happening because in the result I gave, $$\log\left(\frac{4}{5}\right)=\log(4)-\log(5)=-(\log(5)-\log(4))=-\log\left(\frac{5}{4}\right)$$. Thanks from uint Last edited by ZardoZ; November 28th, 2014 at 07:03 PM. November 28th, 2014, 11:53 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 You should have used a definite integral. Thanks from uint Tags logarithm, partial faction, sum, work ### \infty doesn't work

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