1. Show that the sum to \(n\) terms of the series \[\frac{1}{2.1}-\frac{1}{5.2}-\frac{1}{8.5}-\cdots+ \frac{1}{(3r-1)(4-3r)}\cdots\] is \(\frac{n}{3n-1}\).

Approach 1: Proof by induction

We first check for the case \(n=1\): \[\frac{1}{2\times 1}=\frac{1}{3 \times 1-1}\] which is true, so this holds.

Now we assume the result holds for \(n=k\), that is, suppose we know that \[\sum_{r=1}^k\frac{1}{(3r-1)(4-3r)}=\frac{k}{3k-1}.\] Then we have that \[\begin{align*} \sum_{r=1}^{k+1}\frac{1}{(3r-1)(4-3r)} &=\sum_{r=1}^{k}\frac{1}{(3r-1)(4-3r)}+\frac{1}{(3(k+1)-1)(4-3(k+1))}\\ &=\frac{k}{3k-1}+\frac{1}{(3(k+1)-1)(4-3(k+1))}\\ &=\frac{k}{3k-1}+\frac{1}{(3k+2)(1-3k)} \\ &=\frac{k(3k+2)-1}{(3k-1)(3k+2)}\\ &=\frac{3k^2+2k-1}{(3k-1)(3k+2)}\\ &=\frac{(3k-1)(k+1)}{(3k-1)(3k+2)} \\ &=\frac{k+1}{3k+2}\\ &=\frac{k+1}{3(k+1)-1}. \end{align*}\]

This means that our formula holds for \(k+1\), as desired. (The cancellation we did was permissible since \(3k-1\) is non-zero.)

So our formula holds for \(n = 1\), and if it holds for \(n = k\), it will hold for \(n = k+1\). Therefore, by induction, the required result holds for all positive integers \(n\).

While this proof works, it doesn’t tell us how the result was discovered in the first place. Approach 2 sheds light on this.

Approach 2: Partial fractions

We can rewrite the general term using partial fractions: \[\frac{1}{(3r-1)(4-3r)} = \frac{A}{3r-1} + \frac{B}{4-3r}\] so \[1 = A(4-3r) + B(3r-1).\] Putting \(r=1/3\) gives \(A=1/3\), while putting \(r=4/3\) gives \(B=1/3\) too, so \[\frac{1}{(3r-1)(4-3r)} = \frac{1}{3}\left(\frac{1}{3r-1} + \frac{1}{4-3r}\right).\]

Thus our sum becomes \[\begin{align*} \frac{1}{2.1}-&\frac{1}{5.2}-\cdots+\frac{1}{(3n-1)(4-3n)}\\ &=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{1}+ \frac{1}{5}-\frac{1}{2}+ \cdots + \frac{1}{3n-1}-\frac{1}{3n-4}\right), \end{align*}\]

where we have written \(\dfrac{1}{4-3n}\) as \(-\dfrac{1}{3n-4}\).

We see that the first fraction in each pair (beginning with \(1/2\)) cancels with the second fraction in the following pair, leaving only \[\frac{1}{3}\left(\frac{1}{1}+\frac{1}{3n-1}\right) =\frac{1}{3}\cdot\frac{3n}{3n-1}=\frac{n}{3n-1}\] as required.

  1. Given that for \[-1<x<+1, \quad \log_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}- \quad\text{to infinity,}\] deduce the sum to infinity of \[x+\frac{x^5}{5}+\frac{x^7}{7}+\cdots.\]

The first thing to notice is that the series we’d like to sum contains some of the same terms as the expression for \(\log_e(1+x)\).

We’d like to remove the terms of the form \(\dfrac{x^{2n}}{2n}\), as well as the \(\dfrac{x^3}{3}\).

Approach 1

What happens if we consider \(\log_e(1+x^2)\) instead? Replacing \(x\) by \(x^2\) in the given formula, we find that \[\log_e(1+x^2)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\cdots\] which is nearly what we want – but to cancel out the terms we wanted we need these all to have the same sign and their coefficients need the be half their current size. Let’s try \[\begin{align*} \frac{1}{2}\log_e(1-x^2)&=-\frac{x^2}{2}-\frac{(-x^2)^2}{4}+ \frac{(-x^2)^3}{6}-\cdots\\ &=-\frac{x^2}{2}-\frac{x^4}{4}-\frac{x^6}{6}-\cdots \end{align*}\] so we have that \[\begin{align*} x+\frac{x^5}{5}+\frac{x^7}{7}+... &=\log_e(1+x)-\frac{1}{2}\log_e(1-x^2)-\frac{x^3}{3}\\ &=\log_e\left(\frac{1+x}{\sqrt{1-x^2}}\right)-\frac{x^3}{3}\\ &=\frac{1}{2}\log_e\left(\frac{(1+x)^2}{1-x^2}\right)-\frac{x^3}{3}\\ &=\frac{1}{2}\log_e\left(\frac{1+x}{1-x}\right)-\frac{x^3}{3} \end{align*}\]

Approach 2

To cancel out the even terms in the expression for \(\log_e(1+x)\), we can first negate all of the odd terms by replacing \(x\) by \(-x\): \[\log_e(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots\] and then subtract this from \(\log_e(1+x)\) to cancel the even terms, giving \[\log_e(1+x)-\log_e(1-x)=2x+\frac{2x^3}{3}+\frac{2x^5}{5}+\cdots\] If we now divide this by \(2\) and subtract the \(x^3\) term, we deduce our required result: \[x+\frac{x^5}{5}+\frac{x^7}{7}+\cdots= \frac{1}{2}(\log_e(1+x)-\log_e(1-x))-\frac{x^3}{3},\] as before.