General Term of a given series , Where $\sum^{n}_{r=1}U_r=\frac{3n}{2n+1}$ General term $U_r $ of a given series , Where $$\sum^{n}_{r=1}U_r=\frac{3n}{2n+1}$$
I can evaluate that $$ U_1=\frac{3}{3}$$
$$ U_2=\frac{1}{5}$$
$$ U_3=\frac{3}{35}$$
$$ U_4=\frac{1}{21}$$
$$ U_5=\frac{1}{33}$$
How can I recognize a pattern ? 
Yes I can guess a pattern ! But how can I mathematically prove it ?
 A: Let $S_n=\frac{3n}{2n+1}$.  Then $U_n=S_n-S_{n-1}$
This is because
$$\sum_{r=1}^{n} U_r - \sum_{r=1}^{n-1} U_r=$$ $$\left(U_1+U_2+\ldots+U_{n-1}+U_n\right)-\left(U_1+U_2+\ldots+U_{n-1}\right)=U_n$$
and as you said, $U_1=1$
A: When trying to spot a pattern in a problem like this, it’s often helpful not to reduce the fractions. In this case every numerator is initially $3$, and the fractions are:
$$\begin{align*}
U_1&=\frac33\\
U_2&=\frac3{15}\\
U_3&=\frac3{35}\\
U_4&=\frac3{63}\\
U_5&=\frac3{99}\;.
\end{align*}$$
Since all of the numerators are $3$, we can concentrate on the denominators. These are:
$$3,15,35,63,99$$
You might notice that each is $1$ less than perfect square. If you do, you’ll quickly conjecture that 
$$U_n=\frac3{(2n)^2-1}=\frac3{4n^2-1}\;.$$
Alternatively, you might try factoring them:
$$\begin{align*}
3&=1\cdot3\\
15&=3\cdot5\\
35&=5\cdot7\\
63&=7\cdot9\\
99&=9\cdot11
\end{align*}$$
This would suggest that
$$U_n=\frac3{(2n-1)(2n+1)}=\frac3{4n^2-1}\;.$$
In either case you can then prove the result by induction on $n$.
Of course if you happen to see JasonM’s approach, you bypass the pattern-recognition stage and need no induction argument.
