The denominator isn't consecutive

$\text{Just the common method of partial fractions.}\\ \displaystyle f(n)=\sum_1^{\infty} \dfrac{n+2}{n*(n+1)*(n+3)} \\$ $\therefore ~ \dfrac{n+2}{n*(n+1)*(n+3)}= \dfrac A n + \dfrac B{n+1} + ~ \dfrac C{n+3} \\$ $Solving,~~ \dfrac{n+2}{n*(n+1)*(n+3)}= \frac 2 3*\dfrac 1 n - \frac 1 2 *\dfrac 1 {n+1} - \frac 1 6 * \dfrac 1 {n+3}\\ \therefore ~\displaystyle f(n)=\sum_1^{\infty} \left \{ \frac 2 3*\dfrac 1 n - \frac 1 2 *\dfrac 1 {n+1} - \frac 1 6 * \dfrac 1 {n+3} \right \} \\ \therefore ~\displaystyle f(n)= \frac 2 3*\sum_1^3 \dfrac 1 n+ \frac 2 3*\sum_1^{\infty} \dfrac 1 {n+3} - \frac 1 2 *sum_1^2 \dfrac 1 {n+1} - \frac 1 2 *\sum_1^{\infty} \dfrac 1 {n+1} - \frac 1 6 *\sum_1^{\infty} \dfrac 1 {n+3} \\ \frac 2 3*(\frac 1 1+ \frac 1 2+ \frac 1 3) - \frac 1 2 *(\frac 1 2+ \frac 1 3)+0+0+...\\ \dfrac {22}{18} -\dfrac 5{12}=\dfrac {29}{36}. ~~~~~~a+b=29+36= \Huge ~~\color{#D61F06}{65}.$

Note: The first term 'n' to become 'n+3' needs to move 3 steps so limit 1 to 3. Similarly the second term, to move from 'n+1' to 'n+3' needs to move 2 steps so limit 1 to 2. Since the limit is infinity, all end at infinity. If the limit was say L, term 'n' would end at L - 3, and 'n+1' would end at L-2.

General Term Of The Series Is

T = r+2 / r(r+1)(r+3)

T = 1/2 (r+1+r+3/r(r+1)(r+3)

T = 1/2( 1/r(r+3) + 1/r(r+1) )

We Break Down Series Into Two Telescoping Series

T2 = 1/r(r+1)

Its A Very Very Trivial Telescoping Series

T2 = 1/1 - 1/2 + 1/2 -1/3 + 1/3 -1/4.........

as limit is tending towards infinity so we T2 = 1

Main Thing Is To Evaluate T1 = 1/r(r+3)

Now We Use A Good Trick Here

T1 = 1/3 (1/r - 1/r+1) + 1/3(1/r+1 - 1/r+2) + 1/3(1/r+2-1/r+3)

These Three All Are Telescoping Series Individually

The Sum Evaluates To

T1 = 1/3 (1+1/2+1/3)

Final Sum => 1/2(T1+T2) = 29/36

I Thought Of This Trick While Solving This Problem Only !! Keep Posting

$\frac{3}{1 \times 2 \times 4}+\frac{4}{2 \times 3 \times 5}+\frac{5}{3 \times 4 \times 6}+...$ $=\sum_{n=3}^\infty\frac{n}{(n-2)(n-1)(n+1)}=\sum_{n=3}^\infty\frac{4}{6n-12}-\frac{3}{6n-6}-\frac{1}{6n+6}$ $=\frac{4}{6}-\frac{3}{12}-\frac{1}{24}+\frac{4}{12}-\frac{3}{18}-\frac{1}{30}+\frac{4}{18}-\frac{3}{24}-\frac{1}{36}+\frac{4}{24}...$ $=\frac{4}{6}+\frac{1}{12}+\frac{1}{18}=\frac{24+3+2}{36}=\frac{29}{36}$ $29+36=\boxed{65}$