An awkward sum

I’m not sure where I got this question from. It asks for the value of the infinite sum:

$\frac{3}{1\times 2\times 3}+\frac{5}{2\times 3\times 4}+\frac{7}{3\times 4\times 5}+\dots$

Have a go if you’d like; spoilers are below the line.

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The first thing I would do is work out the general term. The numerators are going up by two each time, so I reckon it’s $\frac{2k+1}{k(k+1)(k+2)}$, and we want

$S=\sum _{k=1}^{\mathrm{\infty }}\frac{2k+1}{k(k+1)k+2}$.

How do we get that? Partial fractions look like they might be our friend.

It turns out that it’s a bit less messy to work out $2S$.

• $\frac{4k+2}{k(k+1)(k+2)}=\frac{A}{k}+\frac{B}{k+1}+\frac{C}{k+2}$
• $4k+2=A(k+1)(k+2)+Bk(k+2)+Ck(k+1)$ [*]

Aside: let’s solve this two ways

First way: simultaneous equations. I don’t much care for this way, but it seems that some do.

Matching coefficients:

• $[k^2]$: $0=A+B+C$
• $[k]:4=3A+2B+C$
• $[1]:2=2A$

This gives us $A=1$ immediately, with $B=2$ and $C=-3$ close behind.

Alternatively, we can substitute any numbers we like into [*] and get:

• $[k=0]$: $2=A$
• $[k=-1]:-2=-B$, so $B=2$
• $[k=-2]:-6=2C$, so $C=-3$, as before.

Back to the question at hand

So, $2S=\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{k}+\frac{2}{k+1}-\frac{3}{k+2})$.

Let’s write out a few terms of that:

k
1	$\frac{1}{1}$	$\frac{2}{2}$	$-\frac{3}{3}$
2	.	$\frac{1}{2}$	$\frac{2}{3}$	$-\frac{3}{4}$
3	.	.	$\frac{1}{3}$	$\frac{2}{4}$	$-\frac{3}{5}$

It’s hopefully quite clear that the terms with a denominator of 3 sum to zero, and so will the terms with a denominator of 4, 5 and so on.

All we’re left with are the denominators of 1 (which sum to 1) and 2 (which sum to $\frac{3}{2}$), so $2S=\frac{5}{2}$ and $S=\frac{5}{4}$.

Alternatively

We could use generating functions. It’s massively overkill, but it’s a good excuse to use the technique. Let’s imagine that our sum is in fact a function of $x$, $S(x)=\sum _{k=1}^{\mathrm{\infty }}\frac{2k+1}{k(k+1)k+2}{x}^k$.

We can reuse our partial fractions and rewrite that as $2S(x)=\sum _{k=1}^{\mathrm{\infty }}(\frac{x^k}{k}+\frac{2x^k}{k+1}-\frac{3x^k}{k+2})$

And what we have there are three logarithmic series. Let’s examin them in turn:

• $\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k}=\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\dots$

• $\cdots =-\ln (1-x)$.

• $\sum _{k=1}^{\mathrm{\infty }}\frac{2x^k}{k+1}=2(\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\dots )$

• $\cdots =-\frac{2}{x}(\ln (1-x)+x)$.

• $\sum _{k=1}^{\mathrm{\infty }}-\frac{3x^k}{k+2}=-3(\frac{x}{3}+\frac{x^2}{4}+\frac{x^3}{5}\dots )$

This one’s a bit trickier. I would start with $-\ln (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots$ and think “I want to get rid of the first couple of terms and divide by $x^2$”, giving me…

• $\cdots =3\left(\frac{\ln (1-x)+x+\frac{x^2}{2}}{x^2}\right)$
• or more nicely, $\frac{3}{2x^2}(2\ln (1-x)+2x+x^2)$

So, altogether, $2S(x)=\frac{3}{2x^2}(2\ln (1-x)+2x+x^2)-\frac{2}{x}(\ln (1-x)+x)-\ln (1-x)$.

That screams “regroup!” at me. I’m going to turn the whole thing into a single fraction:

• $2S(x)=\frac{(6\ln (1-x)+6x+3x^2)-(4x\ln (1-x)+4x)-2x^2\ln (1-x)}{2x^2}$
• $\cdots =\frac{(6-4x-2x^2)\ln (1-x)+3x^2+2x}{2x^2}$

We’re interested in the value of this when $x=1$, although there’s a little technical difficulty: this is undefined when $x=1$, because $\ln (0)$ is not a number.

Fortunately, the factor in front of the logarithm also evaluater to zero, and $\underset{z\to 0}{lim}z\ln (z)=0$ ¹

So, in the limit, we’re left with $2S(1)\to \frac{3+2}{2}$, so $S(1)\to \frac{5}{4}$ as before.

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I’m absolutely not saying that the generating functions way is easier, but the technique is an interesting and wide-ranging one. Try applying it to your favourite difficult sequences today!

Footnotes:

1. Prove this, if you’d like to.