Find the error in an inductive "proof"

Consider the statement:

$A(n): 1 + 2 + \cdots + n = \frac{1}{8} (2n+1)^2.$

Prove that if the statement is true for an integer $k$ , then it is also true for $k+1$ .
What is wrong with the claim that this proves that $A(n)$ is true for all $n$ ?
Make a change to turn this into a true statement and prove it for all positive integers.

Proof. Assume the statement is true for an integer $k$ . Then, adding $k+1$ to both sides of the formula we assume is true for the case $k$ ,

$\begin{align*} 1+2+\cdots + k + (k+1) &= \frac{1}{8} (2k+1)^2 + (k+1) & (\text{Ind. Hyp.}) \\ &= \frac{k^2}{2} + \frac{k}{2} + \frak{1}{8} + k + 1 \\ &= \frac{4k^2 + 12k + 9}{8} \\ &= \frac{1}{8} (2(k+1)+1)^2 \end{align*}$

Thus, the statement is true for $k+1$ if it is true for $k. \qquad \blacksquare$
Of course, the above is not a proof by induction, and does not show that the statement is true for all $n$ since we have not shown that it is true for any base case. We have shown that if it is true for some $n$ , then it is true for $n+1$ . To show that it is true for all $n$ , we must show that it is true for the case $n=1$ (or some other starting point). As one can calculate, it is not true for $n=1$ , or any other $n$ for that matter.
In the case $n=1$ we have 1 on the left, and $\frac{9}{8}$ on the right. So, we make the following:
Claim:

$A(n): 1 + 2 + \cdots + n < \frac{1}{8} (2n+1)^2.$

Proof. We have already shown it is true for $n=1$ since $1 < \frac{9}{8}$ . Now assume it is true for some integer $k$ , then using the inductive hypothesis and adding $k+1$ to both sides we have

$\begin{align*} 1 +2 + \cdots + k + (k+1) &< \frac{1}{8} (2k+1)^2 + (k+1) \\ &< \frac{1}{8} (4k^2 + 12k + 9) \\ &< \frac{1}{8} (2(k+1)+1)^2. \end{align*}$

Thus, the inequality is true for all $n. \qquad \blacksquare$

Stumbling Robot