Review question

# Can we show the sum of this series to $n$ terms is $n/(3n-1)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5634

## Solution

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.