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We are asked to evaluate the following sum

$$\sum_{r=1}^\infty {r\over (r+1)!}$$

I thought it best to write the expression out as a difference and use the method of differences to find the sum.

However, how would one go about converting it into a difference? I lucked upon the solution, as it turns out:

$$ {r\over (r+1)!} \equiv {1 \over r!} - {1\over (r+1)!} $$

But it was simply a result of me playing about and was not rigorous at all. What would've been a sure-fire method to reach this? More generally how should one approach expression like these when looking to convert them into differences, similar to how converting some expressions like ${1\over 4r^2-1 }$ into partial fractions can ease the solution?

Shaheer ziya
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3 Answers3

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There is a standard trick that involves adding and subtracting the same number (often $1$) to make things line up. In this case $$ \frac{r}{(r+1)!}=\frac{r+1-1}{(r+1)!}\\ =\frac{r+1}{(r+1)!}-\frac{1}{(r+1)!}\\ =\frac{1}{r!}-\frac{1}{(r+1)!} $$ How would one discover this? Well, we would really like to have $r+1$ in the numerator, because that would cancel very nicely with the denominator. So we make it so that we have $r+1$ in the numerator, by adding $1$, and then subtracting $1$. The rest is standard arithmetic manipulations.

Arthur
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Well in case of factorials it is always a good idea to use the method of comparing coefficients.$$\frac{r}{(r+1)!}=\frac {A}{(r)!}-\frac{A}{(r+1))!}$$

The rest is just comparing coefficients.

The pointis to bring it of the form of $|f(r)-f(r+1)|$

In case of $$\frac{1}{4r^2-1}=\frac{1}{4}\frac{1}{(r-1/2)(r+1/2)}=\frac{1}{4}(\frac{A}{r-1/2}-\frac{A}{r+1/2})$$

Sometimes it gets harder:say we have to find $$\sum r(r+1)(r+2)(r+3)$$

In this case we can write $T_r=r(r+1)(r+2)(r+3)$$

$$=\frac{1}{5}r(r+1)(r+2)(r+3)(r+4-(r-1))$$

$$=\frac{1}{5}r(r+1)(r+2)(r+3)(r+4)-(r-1)(r)(r+1)(r+2)(r+3)=V(r-1)-V(r)$$

The summary is in case consecutive coefficients come trying to manipulate a constant term to the form $r+k-r$ sometimes works.

Exercise:Evaluate $$\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}$$

Hint:

use partial fractions $$\frac{1}{r(r+1)(r+2)}=\frac{A}{r}-\frac{B}{r+1}+\frac{C}{r+2}$$

Although some times we have to think a bit differently as in this problem: sum of terms of series

Albus Dumbledore
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I would not say "writing out the expression as a difference", but more "turning it into more familiar terms". There is no general method for this.

In $\dfrac r{(r+1)!}$ we don't see a simplification, because we need $r+1$ at the numerator. So we can use an artifice and write

$$\dfrac r{(r+1)!}=\dfrac{r+1-1}{(r+1)!}=\frac1{r!}-\frac1{(r+1)!}$$ and voilà, by chance. We now can work out the partial $e$ series.