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November 28th, 2014, 01:23 AM   #1
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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!
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November 28th, 2014, 08:21 AM   #2
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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.
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November 28th, 2014, 02:15 PM   #3
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Great, thanks!
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November 28th, 2014, 07:00 PM   #4
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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.
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November 28th, 2014, 11:53 PM   #5
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You should have used a definite integral.
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