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 ?

• If you think to have recognised a pattern, then the easiest way to check if it holds is usually by induction. Jul 16 '16 at 7:32

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, $U_1=1$

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.