Evaluating the Sum of $ \frac {1} {3}+\frac {1} {15}+\frac {1} {35}+\ldots +\frac {1} {4n^{2}-1} $ How would you evaluate the sum of this sequence ?
$$
\dfrac {1} {3}+\dfrac {1} {15}+\dfrac {1} {35}+\ldots +\dfrac {1} {4n^{2}-1}
$$
I realise the expression can be factorised but I can't really see what this can tell you.
 A: $$\sum_{n=1}^{\infty }\frac{1}{4n^2-1}=\sum_{n=1}^{\infty }\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$$
$$=\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....\right]=\frac{1}{2}$$
A: Hint
$$\frac{1}{4n^2-1} = \frac{1/2}{2n-1} - \frac{1/2}{2n+1}$$
A: As shown in previous answers, you face a telescopic sum since $$\frac{1}{4i^2-1}= \frac 12\Big(\frac{1}{2i-1} - \frac{1}{2i+1}\Big)$$ So if $$S_n=\sum_{i=0}^n \frac{1}{4i^2-1}$$ $$2S_n=\sum_{i=0}^n\frac{1}{2i-1}-\sum_{i=0}^n\frac{1}{2i+1}$$ $$2S_n=\Big(\frac 11+\frac 13+\frac 15+\cdots+\frac 1{2n-1}\Big)-\Big(\frac 13+\frac 15+\frac 17+\cdots+\frac 1{2n-1}+\frac 1{2n+1}\Big)$$ So, after elimination $$2S_n=1-\frac 1{2n+1}=\frac {2n}{2n+1}$$ $$S_n=\frac {n}{2n+1}$$ If $n$ is large, you could perform the long division and get $$S_n\approx \frac{1}{2}-\frac{1}{4 n}+\cdots$$ which shows the limit and also how it is approached.
A: $$\sum_{r=1}^n \frac 1{4r^2-1}=\sum_{r=1}^n \left(\frac r{2r+1}-\frac{r-1}{2r-1}\right)=\frac n{2n+1}\qquad\blacksquare$$
