Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$

Question

$\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$

The general term seems to be $$T_r= \frac{r+2}{r(r+1)(r+3)}.$$

I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.

Do I have to use something else? Or am I missing some easy manipulation?

Answer 1

\begin{align*} T_r &= \frac{1}{r(r+1)} - \frac{1}{r(r+1)(r+3)} \\ &= \frac{1}{r(r+1)}- \left(\frac{1}{2}\cdot \frac{(r+3)-(r+1)}{r(r+1)(r+3)}\right) \\ &= \frac{1}{r}-\frac{1}{r+1}- \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot \frac{1}{r(r+3)} \\ &= \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot\frac{1}{3}\left(\frac{1}{r}-\frac{1}{r+3}\right) \\ &= \frac{2}{3}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{6}\left(\frac{1}{r+1}-\frac{1}{r+2}\right)+ \frac{1}{6}\left(\frac{1}{r+2}-\frac{1}{r+3}\right) \end{align*} You can now see telescope and quicky add the terms.

Answer 2

Hint:

$$\frac{r+2}{r(r+1)(r+3)} = \frac2{3r}-\frac 1{2(r+1)} - \frac1{6(r+3)}$$

Answer 3

We have that

$$a_r=\frac{r + 2}{r (r + 1) (r + 3)} = \frac2{3 r}-\frac1{2 (r + 1)} - \frac1{6 (r + 3)}=$$$$= \frac1{2r}-\frac1{2 (r + 1)}+ \frac1{6r} - \frac1{6 (r + 3)}=$$

$$= \frac12\left(\frac1{r}-\frac1{r + 1}\right)+ \frac16\left(\frac1{r} - \frac1{r + 3}\right)$$

and therefore

$$\sum_1^n a_r=\frac12 \sum_1^n \left(\frac1{r}-\frac1{r + 1}\right)+\frac16\sum_1^n\left(\frac1{r} - \frac1{r + 3}\right)=$$

$$=\frac12\left(1-\frac1{n+1}\right)+\frac16\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right)=$$

$$=\frac{29}{36}-\frac2{3(n+1)}-\frac1{6(n+2)}-\frac1{6(n+3)}=$$

$$=\frac{29 n^3+138 n^2+157 n}{36(n+1)(n+2(n+3)}$$

See also the related Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$.

Answer 4

There still is the brutal way:

$$ \frac{r+2}{r(r+1)(r+3)}\stackrel{\text{PFD}}{=}\frac{2}{3}\cdot\frac{1}{r}-\frac{1}{2}\cdot\frac{1}{r+1}-\frac{1}{6}\cdot\frac{1}{r+3} \tag{1}$$ leads to: $$ \frac{r+2}{r(r+1)(r+3)} = \frac{1}{6}\int_{0}^{1} x^{r-1}\left(4-3x-x^3\right)\,dx \tag{2}$$ then by summing both sides over $r\geq 1$: $$ \sum_{r\geq 1}\frac{r+2}{r(r+1)(r+3)}=\frac{1}{6}\int_{0}^{1}\frac{4-3x-x^3}{1-x}\,dx=\frac{1}{6}\int_{0}^{1}\left(4+x+x^2\right)\,dx \tag{3} $$ and simplifying: $$ \sum_{r\geq 1}\frac{r+2}{r(r+1)(r+3)}=\frac{1}{6}\left(4+\frac{1}{2}+\frac{1}{3}\right)=\color{red}{\frac{29}{36}}.\tag{4}$$ The partial sums are more directly computed via $(1)$: $$\begin{eqnarray*} \sum_{r=1}^{n}\frac{r+2}{r(r+1)(r+3)}&=&\frac{2}{3}H_n-\frac{1}{2}\left(H_{n+1}-1\right)-\frac{1}{6}\left(H_{n+3}-\frac{11}{6}\right)\\&=&\frac{157 n+138 n^2+29 n^3}{36 (n+1)(n+2)(n+3)}.\tag{5} \end{eqnarray*}$$

Answer 5

I copy this from my answer to a duplicate problem:

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.

On page 316, Hall & Knight 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.

Answer 6

HINT:

Using Partial Fraction Decomposition,

$$\frac {r+2}{r(r+1)(r+3)}=\frac Ar+\frac B{r+1}+\frac C{r+3}$$ wherere $A,B,C$ are arbitrary constants

$$\implies r+2=A(r+1)(r+3)+Br(r+3)+Cr(r+1)$$

We can put $r=0,-1,-3$

or we can compare the coefficients of the different powers of $r$

to determine $A,B,C$