Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ 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 $1/2$?
 A: 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,
$$
\sum_{k=1}^\infty\frac{1}{(2k-1)(2k+1)}=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left(\frac{1}{2}-\frac{1}{2(2n+1)}\right)=\frac{1}{2}
$$
A: 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
$$\begin{eqnarray*}
\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)  &=&\lim_{N\rightarrow
\infty }\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)  \\
&=&\lim_{N\rightarrow \infty }\left( a_{1}-a_{N+1}\right)  \\
&=&a_{1}-\lim_{N\rightarrow \infty }a_{N+1} \\
&=&a_{1}-\lim_{N\rightarrow \infty }a_{N} \\
&=&a_{1}-\lim_{n\rightarrow \infty }a_{n}.\end{eqnarray*}$$
A: 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%)
A: 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. $$S=\frac{1}{2}$$
