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Show that $$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$

I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox. If the series continues infinitely then each term is just going to get smaller and smaller.

Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$?

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7  
The magic words are "telescoping series". –  David Mitra Jul 31 '12 at 23:17
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Thank you for the magic words, I really just needed a push in the right direction! –  Jack Thompson Jul 31 '12 at 23:19
    
@JackThompson, co may be you should write your own an answer to this question? –  Norbert Jul 31 '12 at 23:20
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The magic words are a push in the right direction. You are supposed to take the hint to look up the magic words and see how to use what you find to solve the problem. Or maybe you did that - it's not clear to me from what you've written. –  Gerry Myerson Aug 1 '12 at 1:55
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4 Answers

Solution as per David Mitra's hint in a comment.


Write the given series as a telescoping series and evaluate its sum:

$$\begin{eqnarray*} S &=&\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots \\ &=&\sum_{n=1}^{\infty }\frac{1}{\left( 2n-1\right) \left( 2n+1\right) } \\ &=&\sum_{n=1}^{\infty }\left( \frac{1}{2\left( 2n-1\right) }-\frac{1}{ 2\left( 2n+1\right) }\right)\quad\text{Partial fractions decomposition} \\ &=&\frac{1}{2}\sum_{n=1}^{\infty }\left( \frac{1}{2n-1}-\frac{1}{2n+1} \right) \qquad \text{Telescoping series} \\ &=&\frac{1}{2}\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right), \qquad a_{n}= \frac{1}{2n-1},a_{n+1}=\frac{1}{2\left( n+1\right) -1}=\frac{1}{2n+1} \\ &=&\frac{1}{2}\left( a_{1}-\lim_{n\rightarrow \infty }a_{n}\right) \qquad\text{see below} \\ &=&\frac{1}{2}\left( \frac{1}{2\cdot 1-1}-\lim_{n\rightarrow \infty }\frac{1 }{2n-1}\right) \\ &=&\frac{1}{2}\left( 1-0\right) \\ &=&\frac{1}{2}. \end{eqnarray*}$$

Added: The sum of the telescoping series $\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)$ is the limit of the telescoping sum $\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) $ as $N$ tends to $\infty$. Since

$$\begin{eqnarray*} \sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) &=&\left( a_{1}-a_{2}\right) +\left( a_{2}-a_{3}\right) +\ldots +\left( a_{N-1}-a_{N}\right) +\left( a_{N}-a_{N+1}\right) \\ &=&a_{1}-a_{2}+a_{2}-a_{3}+\ldots +a_{N-1}-a_{N}+a_{N}-a_{N+1} \\ &=&a_{1}-a_{N+1}, \end{eqnarray*}$$

we have

$$\begin{eqnarray*} \sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right) &=&\lim_{N\rightarrow \infty }\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) \\ &=&\lim_{N\rightarrow \infty }\left( a_{1}-a_{N+1}\right) \\ &=&a_{1}-\lim_{N\rightarrow \infty }a_{N+1} \\ &=&a_{1}-\lim_{N\rightarrow \infty }a_{N} \\ &=&a_{1}-\lim_{n\rightarrow \infty }a_{n}.\end{eqnarray*}$$

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2  
Not leaving much for OP to do. –  Gerry Myerson Aug 1 '12 at 1:57
    
As @Gerry said. Schade. –  Did Aug 1 '12 at 6:20
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@GerryMyerson I agree. I will take your words into account in future answers. –  Américo Tavares Aug 1 '12 at 8:16
    
@did I agree. I will take your words into account in future answers. –  Américo Tavares Aug 1 '12 at 8:17
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You can prove it with partial sums: $$ S_n=\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n\left(\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\right)=\frac{1}{2}\left(\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=2}^{n+1}\frac{1}{2k-1}\right) $$ $$ =\frac{1}{2}\left(\frac{1}{2(1)-1}-\frac{1}{2(n+1)-1}\right)=\frac{1}{2}-\frac{1}{2(2n+1)} $$ Hence, $$ \sum_{k=1}^\infty\frac{1}{(2k-1)(2k+1)}=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left(\frac{1}{2}-\frac{1}{2(2n+1)}\right)=\frac{1}{2} $$

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Here is an intuitive way to image this problem.
Imagine a teacher gives you a test with an odd number of problems. You get the first problem incorrect. Then with the next two problems, you miss one and get one right. As the number of problems increases to infinity, your score will approach 1/2( or 50 %)

1 problem test , your score is 0/1
3 problem test , your score is 1/3
The difference bewtween 1/3 and 0/1 is 1/3
5 problem test , your score is 2/5
The difference between 2/5 and 1/3 is 1/15
7 problem test , your score is 3/7
The difference between 3/7 and 2/5 is 1/35

(2n-1) problem test , your score is (n-1)/(2n-1)
(2n+1) problem test, your score is (n)/(2n+1)

The difference of your score between a (2n+1) test and a (2n-1) test is:
(n+1)/(2n+1) - (n)/(2n-1) which simplifies to
1/(2n-1)(2n+1)

The sum of this series (starting with n =1 ) goes to 1/2

PS, you can use this idea to prove that 1/2 + 1/6 +1/12 +1/20 +... goes to 1
In this example, you miss the first problem on the test and then get the rest correct. As the number of problems goes to infinity, your score approaches 1( or 100%)

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consider,$$f(n)=\frac{1}{(2n-1) \cdot (2n-1+2)}$$

where $n$ is a natural number

$$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty\frac{1}{(2n-1) \cdot (2n+1)}$$

Let, $$\sum_{n=1}^{\infty} f(n) = S$$

i.e. $$S=\sum_{n=1}^\infty\frac{1}{(2n-1)\cdot(2n+1)}$$ i.e. $$S=\sum_{n=1}^\infty(\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})$$

i.e. $$S=(\frac{1}{2})\left(\sum_{n=1}^\infty(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$

i.e. $$S=\lim_{n \to \infty}\left((\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{2})(\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$

i.e. $$S=\lim_{n \to \infty}(\frac{1}{2})\left((\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$

i.e. $$S=\lim_{n \to \infty}(\frac{1}{2})\left(\frac{1}{1} - \frac{1}{2n+1}\right)$$

i.e. $$S=(\frac{1}{2})\left(\frac{1}{1} - \lim_{n \to \infty}(\frac{1}{2n+1})\right)$$

i.e. $$S=(\frac{1}{2})\left(\frac{1}{1} - 0\right)$$

i.e. $$S=(\frac{1}{2})\left(\frac{1}{1}\right)$$

i.e. $$S=\frac{1}{2}$$

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3  
I think you want $f(n)=\frac 1{(2n-1)(2n+1)}$ to make $S$ the sum of it. Then the first line starting with i.e. is incorrect, as it would include $\frac 1{2\cdot 4}$ and others with even numbers in the denominators. This gets cancelled in the third line starting with i.e. –  Ross Millikan Aug 1 '12 at 3:32
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Aside from the mistakes, what does this add to the earlier answer from Americo? –  Gerry Myerson Aug 1 '12 at 7:23
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Notation: You mean $$\sum_{n=1,3,\dots}^\infty f(n)$$ instead of $$\sum_{n=1,3}^\infty f(n)$$ –  Américo Tavares Aug 2 '12 at 15:16
    
you are right. odd and even numbers both tend to infinity. here odd numbers are considered. –  Rajesh K Singh Aug 3 '12 at 3:40
    
the latest edition is as per convention –  Rajesh K Singh Aug 4 '12 at 17:01
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