Series

Mathematical induction can be used to prove that a series is equivalent to some expression. In general, we follow these basic steps :

Prove for the base case.
Assume for $k$ .
Prove the hypothesis by rearranging the left hand side such that it is equal to the right hand side.

Example

infoNote

Prove the following by induction for $n \in \mathbb{N}$ .

1+2+3+...+n=\frac{n(n+1)}{2}

First, prove this proposition is true for the base case $n=1$ .

\begin{align*} 1&=\frac{(1)((1)+1)}{2} \\\\ &= \frac{2}{2} \\\\ &= 1 \end{align*}

True for base case, $n=1$ .

Next, assume true for some arbitrary number $k$ . Assume true for $n=k$ :

1+2+3+...+k=\frac{k(k+1)}{2}

Finally, prove for $n=k+1$ .

\begin{align*} 1+2+3+...+k+(k+1)&=\frac{(k+1)((k+1)+1)}{2} \\\\ &= \frac{(k+1)(k+2)}{2} \end{align*}

Observe that the series highlighted in red is the same as the inductive hypothesis we assumed is true.

\begin{align*} \textcolor{red}{1+2+3+...+k}+(k+1)&=\frac{(k+1)(k+2)}{2} \end{align*}

Substitute the inductive hypothesis :

\begin{align*} \frac{k(k+1)}{2}+(k+1)&=\frac{(k+1)(k+2)}{2} \end{align*}

The rest involves algebraic manipulation such that LHS is the same as the RHS.

\begin{align*} \frac{k(k+1)}{2}+\frac{(k+1)}{1}&=\frac{(k+1)(k+2)}{2}& \\\\ \frac{k(k+1)}{2}+\frac{2(k+1)}{2}&=& \\\\ \frac{k(k+1)+2(k+1)}{2}&=&\text{\footnotesize\textcolor{gray}{(factor out (\(k+1)\))}} \\\\ \frac{(k+1)(k+2)}{2} &= \end{align*}

LHS = RHS, proven by induction.

Example

infoNote

Prove by induction, the formula for the sum of the first $n$ terms of a geometric series. That is, prove that, for $r \ne 1$ :

a+ar+ar^2+...+ar^{n-1}=\frac{a(1-r^n)}{1-r}

Prove for base case $n=1$ .

\begin{align*} a&=\frac{a(1-r^1)}{1-r} \\\\ &= \frac{a\cancel{(1-r)}}{\cancel{1-r}} \\\\ &= a \end{align*}

True for base case, $n=1$ .

Assume true for $n=k$ :

a+ar+ar^2+...+ar^{k-1}=\frac{a(1-r^k)}{1-r}

Finally, prove for $n=k+1$ .

\begin{align*} a+ar+ar^2+...+ar^{k-1}+ar^{(k+1)-1}&=\frac{a(1-r^{k+1})}{1-r} \\\\ a+ar+ar^2+...+ar^{k-1}+ar^{k}&= \end{align*}

Substitute inductive hypothesis :

\begin{align*} \underbrace{a+ar+ar^2+...+ar^{k-1}}+ar^{k}&=\frac{a(1-r^{k+1})}{1-r} \\\\ \frac{a(1-r^{k})}{1-r} + ar^k&= \\\\ \frac{a(1-r^{k})}{1-r} + \frac{ar^k}{1}&= \\\\ \frac{a(1-r^{k})}{1-r} + \frac{ar^k(1-r)}{1(1-r)}&= \\\\ \frac{a(1-r^k)+ar^k(1-r)}{(1-r)} &= \\\\ \frac{a-ar^k+ar^k-ar^{k-1}}{(1-r)} &= \\\\ \frac{a\cancel{-ar^k}+\cancel{ar^k}-ar^{k-1}}{(1-r)} &= \\\\ \frac{a-ar^{k-1}}{(1-r)} &= \\\\ \frac{a(1-r^{k+1})}{1-r} &= \\\\ \end{align*}

LHS = RHS, proven by induction.

Series Simplified Revision Notes for Leaving Cert Mathematics