Partial fraction decomposition of $(\frac{1}{n} + \frac{-1}{n+1})^p$

Question

Context: I was trying to prove $\;\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3} = 10 - \pi^2$

$\displaystyle \frac{1}{n^3(n+1)^3} = \left( \frac{1}{n} + \frac{-1}{n+1} \right) ^3 $

Partial fraction decomposition of third power is easy, but not p-th power in general.

Below is the proof I am asking for help

I noticed a pattern, for positve integer powers p:

$\displaystyle \left( \frac{1}{n} + \frac{-1}{n+1} \right) ^p = \sum_{k=0}^{p-1}\;(-1)^k \binom{p-1+k}{p-1} \left( \left(\frac{1}{n}\right)^{p-k} + \left(\frac{-1}{n+1}\right)^{p-k} \right)$

I had tried to prove this by induction, but failed.
It turned messy, into sum of sums.

Any ideas?

Answer 1

You want to find the partial fraction expansion for $(\frac1n-\frac1{n+1})^p$, which is the same as $$\frac1{n^p(n+1)^p}.$$

The two roots of the denominator are $n=0$ and $n=-1$, both with multiplicity $p$, so the fundamental theorem for partial fractions tells us that there exist constants $A_1,\dots,A_{p}$ and $B_1,\dots,B_p$ for which $$ \frac1{n^p(n+1)^p}= \frac{A_0}{n^p}+\frac{A_1}{n^{p-1}}+\dots+\frac{A_{p-1}}{n} +\frac{B_0}{(n+1)^p}+\frac{B_1}{(n+1)^{p-1}}+\dots+\frac{B_{p-1}}{n+1} $$ Fix a number $i\in \{1,\dots,p\}$, and let us find $A_i$. To do this, first we multiply the above by $n^p$ to clear all factors of $n$ from all denominators. The result is $$ (n+1)^{-p}=A_0+\dots+A_in^i+\dots+A_{p-1}n^{p-1}+n^p\cdot (\text{stuff with $B$}) $$ Let us think of $n$ as a continuous variable. The trick is to differentiate that equation $i$ times with respect to $n$, and then substitute $0$ for $n$ into the result. Since the constants $A_1,\dots,A_{i-1}$ are attached to powers of $n$ which are less than $i$, the differentiation will make them vanish. Furthermore, everything else besides $A_i$, including the $B$ stuff, will still have a nontrivial power of $n$ after the differentiation, so everything else disappears after setting $n=0$. We get $$ \frac{\mathrm{d}^{i}}{\mathrm{d}n^{i}}(n+1)^{-p}=(i)(i-1)\cdots 2\cdot 1\cdot A_i, $$ It looks weird to differentiate with respect to $n$, so I will use $x$ as an alias for $n$. We have found our formula for $A_i$: $$ A_i=\frac{1}{i!}\frac{\mathrm{d}^{i}}{\mathrm{d}x^{i}}(x+1)^{-p}\Bigg|_{x=0} $$ This formula is well known, so we really could have skipped all of the preceding discussion and jumped straight to here.

This is equivalent to saying that $A_i$ is the coefficient of $x^{i}$ when $(x+1)^{-p}$ is expanded as a Taylor series at $x=0$. Fortunately, we can find this Taylor series expansion using Newton's binomial theorem: $$ (x+1)^{-p}=\sum_{i\ge 0} \binom{-p}{i} x^i=\sum_{i\ge 0} \binom{p+i-1}{p-1}(-1)^ix^i $$ That is, we have found $A_i=(-1)^i\binom{p+i-1}{i-1}$, which exactly agrees with what you wanted to prove.

In the same vein, we can find $$ \begin{align} B_i &=\frac{1}{i!}\frac{\mathrm{d}^{i}}{\mathrm{d}x^{i}}x^{-p}\Bigg|_{x=-1}\\ &=\frac1{i!}\left(((-p)(-p-1)\cdots (-p-i-1)x^{-p-i}\right)\Bigg|_{x=-1}\\ &=(-1)^i\frac{(p+i-1)!}{i!(p-1)!}(-1)^{p+i}\\ &=(-1)^p\binom{p+i-1}{p-1} \end{align} $$

Answer 2

Using just

$$\frac1{n(n+1)} = \frac1n - \frac1{n+1}$$

we can derive the complete decomposition. We have

$$\begin{align*} \frac1{n^3(n+1)^3} &= \left(\frac1n - \frac1{n+1}\right)^3 \\[1ex] &= \frac1{n^3} - \frac3{n^2(n+1)} + \frac3{n(n+1)^2} - \frac1{(n+1)^3} \\[2ex] \frac1{n^2(n+1)} - \frac1{n(n+1)^2} &= \frac1{n^2(n+1)^2} \\[1ex] &= \left(\frac1n - \frac1{n+1}\right)^2 \\[1ex] &= \frac1{n^2} - \frac2{n(n+1)} + \frac1{(n+1)^2} \end{align*}$$

so that

$$\begin{align*} \frac1{n^3(n+1)^3} &= \frac1{n^3} - 3\left(\frac1{n^2} - 2\left(\frac1n - \frac1{n+1}\right) + \frac1{(n+1)^2}\right) - \frac1{(n+1)^3} \\[1ex] &= \frac1{n^3} - \frac3{n^2} + \frac6n - \frac6{n+1} - \frac3{(n+1)^2} - \frac1{(n+1)^3} \end{align*}$$

In the sum, there are two telescoping parts,

$$\sum_{n=1}^\infty \left(\frac1{n^3} - \frac1{(n+1)^3}\right) = 1$$

$$6\sum_{n=1}^\infty \left(\frac1n - \frac1{n+1}\right) = 6$$

and the remaining involves the well-known Basel problem.

$$-3 \sum_{n=1}^\infty \left(\frac1{n^2} + \frac1{(n+1)^2}\right) = -3 \left(\frac{\pi^2}6 + \frac{\pi^2}6 - 1\right) = -\pi^2 + 3$$

Take the total and the result follows.

Answer 3

If the solution via complex integration is acceptable, you can do the following. $$S=\sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3} = \frac{1}{2}\sum_{\underset{n\neq 0,-1}{n=-\infty}}^{\infty} \frac{1}{n^3(n+1)^3}$$ Now let's consider the integral in the complex plane (counter-clockwise) along a big circle with the radius $R\to\infty$ $$I=\int_{C_R}\frac{\pi \cot \pi z}{z^3(z+1)^3}dz$$ the function $\pi\cot\pi z$ has simple poles at $z=0,\pm1, \pm2, ... $ with the residue $=1$. On the one hand, it is straightforward to show that the integral along the circle $\to0$ at $R\to\infty$. On the other hand, it is equal to the sum of residues inside the circle.

Therefore, $$I=2\pi i\sum \operatorname{Res}\frac{\pi \cot \pi z}{z^3(z+1)^3}=2\pi i\Big(\sum_{n=-\infty}^{-2}\frac{1}{n(n+1)}+\sum_{n=1}^{\infty}\frac{1}{n(n+1)}+\underset{z=-1, 0}{\operatorname{Res}}\frac{\pi \cot \pi z}{z^3(z+1)^3}\Big)=0$$ $$2S=-\underset{z=-1, 0}{\operatorname{Res}}\frac{\pi \cot \pi z}{z^3(z+1)^3}\tag{1}$$ In these points we have the poles of order 4. Decomposing the function near $z=0$ $$\frac{\pi \cot \pi z}{z^3(z+1)^3}=\frac{\pi}{z^3}\frac{1-\frac{(\pi z)^2}{2!}+O(z^4)}{\pi z\big(1-\frac{(\pi z)^2}{3!}+O(z^4)\big)}\frac{1}{(1+z)^3}$$ $$=\frac{\pi}{\pi z^4}\Big(1-\frac{(\pi z)^2}{3}+O(z^4)\Big)\Big(1-3z+\frac{3\cdot4}{2!}z^2-\frac{3\cdot4\cdot 5}{3!}z^3+O(z^4)\Big)$$ The residue is the coefficient at $\displaystyle \frac{1}{z}$, and it is equal to $\pi^2-10$ in our case. Exactly in the same way you evaluate the residue at $z=-1$. Taking $z=-1+\epsilon$ $$\frac{\pi \cot \pi z}{z^3(z+1)^3}=-\frac{\pi \cot \pi \epsilon}{\epsilon^3(1-\epsilon)^3}=-\frac{\pi}{\pi \epsilon^4}\Big(1-\frac{(\pi \epsilon)^2}{3}+O(\epsilon^4)\Big)\Big(1+3\epsilon+\frac{3\cdot4}{2!}\epsilon^2-\frac{3\cdot4\cdot 5}{3!}\epsilon^3+O(\epsilon^4)\Big)$$ what gives the same residue: $\displaystyle \pi^2-10$

Putting all in (1) $$2S=-2(\pi^2-10)\,\,\Rightarrow\,\, S=10-\pi^2$$

Answer 4

$\displaystyle \left( \frac{1}{n} + \frac{-1}{n+1} \right) ^p = \sum_{k=0}^{p-1}\;(-1)^k \binom{p-1+k}{p-1} \left( \left(\frac{1}{n}\right)^{p-k} + \left(\frac{-1}{n+1}\right)^{p-k} \right)$

Let $\displaystyle a=\frac{1}{n}, b = \frac{-1}{n+1}$, we have $ab = -(a+b)$

RHS first term is $(a^p+b^p)$

$(a+b)×(a^p + b^p) = (a^{p+1} + b^{p+1}) - (a+b)×(a^{p-1} + b^{p-1})$

It turned into recurrence relation (not just first term, but others too!)
I had stopped there, afriad it turned into big mess.

But, all we need is Pascal's Rule: $\displaystyle\binom{n-1}{k} +\binom{n-1}{k-1} = \binom{n}{k} $

For (p+1)-th power, RHS first term matched, coefficient = $1 = (-1)^0 \binom{p+0}{p}$

Recurrence relation alternating signs matched formula we wished to proof.
Thus, we can ignore sign flips, for now.

For (p+1)-th power, RHS 2nd term coefficient, absolute value, matched.

$\binom{p-1+0}{p-1} + \binom{p-1+1}{p-1} = 1+p = \binom{p+1}{p}$

Pascal's Rule, for (k+1)-th term coefficient, absolute value, also matched.

$\binom{p-1+k}{p} + \binom{p-1+k}{p-1} = \binom{p+k}{p}$

If formula is correct for p-th power, so does for (p+1)-th power.
We also have trivial base case, $p=1\quad ⇒ (a+b)^1 = (a+b)$

By induction, formula is correct.