Find the series' sum: $\frac{3}{1.2.4} + \frac{4}{2.3.5} + \frac{5}{3.4.6}...$ I tried solving it by writing $3$ as $4-1$, $4$ as $5-1$ but it didn't help. I am still studying so I don't know how to solve it by decomposing it into partial fractions.Is there any other method to solve this?
discussion
Formula 203 on page 38 of Jolley, Summation of Series, says $${3\over1\cdot2\cdot4}+{4\over2\cdot3\cdot5}+{5\over3\cdot4\cdot6}+\cdots n{\rm\ terms}={29\over36}-{1\over n+3}-{3\over2(n+2)(n+3)}-{4\over3(n+1)(n+2)(n+3)}$$ from which it is evident that the infinite series converges to $29/36$. Presumably, one can prove the formula by induction, though it may be tedious. Jolley's reference is Hall & Knight, Higher Algebra, London, Macmillan 1899, page 317.
EDIT: I now have access to Hall & Knight. On page 316, they give a general formula for $$S_n=\sum_{k=1}^n{1\over(a+kb)(a+(k+1)b)\cdots(a+(k+r-1)b)}$$ namely, $$S_n=C-{1\over(r-1)b}\cdot{1\over(a+(n+1)b)\cdots(a+(n+r-1)b)}$$ "where $C$ is a quantity independent of $n$, which may be found by ascribing to $n$ some particular value." This doesn't apply directly to the sum we want, but they go $$u_n={n+2\over n(n+1)(n+3)}={(n+2)^2\over n(n+1)(n+2)(n+3)}={n(n+1)+3n+4\over n(n+1)(n+2)(n+3)}$$ and then $$u_n={1\over(n+2)(n+3)}+{3\over(n+1)(n+2)(n+3)}+{4\over n(n+1)(n+2)(n+3)}$$ and now the previous formula applies, giving $$S_n=C-{1\over n+3}-{3\over2(n+2)(n+3)}-{4\over3(n+1)(n+2)(n+3)}$$ Now put $n=1$ to get $C=29/36$ and we're done.