5
$\begingroup$

Evaluate $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$

So in a previous part of the question I calculated that $$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$

So my question is then why does $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)} = -\frac{1}{6}$$

As I though that it would be $\frac{1}{2}$.

$\endgroup$
7
$\begingroup$

Note that\begin{align}\frac{2-r}{r(r+1)(r+2)}&=\frac1r-\frac3{r+1}+\frac2{r+2}\\&=\frac1r-\frac1{r+1}-\frac2{r+1}+\frac2{r+2}.\end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$\frac12-\frac23=-\frac16.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (+1) was a minute late! $\endgroup$ – TheSimpliFire Dec 31 '18 at 15:54
5
$\begingroup$

Note that \begin{align}\sum_{r=2}^\infty\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)&=\sum_{r=2}^\infty\left(\frac1r-\frac1{r+1}\right)-2\sum_{r=2}^\infty\left(\frac1{r+1}-\frac1{r+2}\right)\\&=\left(\frac12-\frac13+\frac13-\frac14+\cdots\right)-2\left(\frac13+\frac14-\frac14+\frac15-\cdots\right)\\&=\frac12-\frac23=-\frac16\end{align}

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ (+1) Believe me: I know what it feels to be a few seconds too late. $\endgroup$ – José Carlos Santos Dec 31 '18 at 15:55
1
$\begingroup$

Actually your finite sum is equal to $$\sum_{r=2}^n\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}-\frac{1}{6}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Let $\dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=\dfrac{am+b}{m(m+1)}$

$\implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$

Set $2b=2\iff b=1,ar=-r\iff a=-1$

$$\implies\sum_{r=1}^n\dfrac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(f(r)-f(r+1)\right)=f(1)-f(n+1)$$

Set $n\to\infty$ and show that $n\to\infty f(n)=0$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

If $H_{n}$ is the $n$-th harmonic number, we have $$\sum_{r=2}^{n}\left(\frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\left(H_{n}-1\right)-3\left(H_{n}-1-\frac{1}{2}\right)+2\left(H_{n}-1-\frac{1}{2}-\frac{1}{3}\right)$$ $$=-1+\frac{3}{2}-\frac{2}{3}=-\frac{1}{6}.$$ Now take the limit as $n\rightarrow+\infty$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

By definition,

$$ \sum_{r=1}^\infty a_r = \lim_{n\to\infty}\sum_{r=1}^n a_r .$$

In your case the above limit is zero. Then:

$$ \sum_{r=2}^\infty a_r = \sum_{r=1}^\infty a_r - a_1 = -a_1 . $$

This reasoning is what @HenriLee was referring to in his answer.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.