<s>[INST] Weird result proving : $1+2+3+...+n<\frac{(2n+1)²}{8} $ 
$$1+2+3+...+n<\frac{(2n+1)²}{8} $$

So, I'm going through Apostol's Calculus, and solving this:

I want to prove the aforementioned inequality, and proceed like this :

*

*$A(1) = 1 < \frac{9}{8}$
$A(2) = 1+2=3 < \frac{25}{8}$
$...$

*

*$A(k)=1+2+3+...+k<\frac{(2k+1)²}{8} | k≥1 $


*$A(k+1)=1+2+3+...+k+(k+1)<\color{red}{\frac{(2(k+1)+1)²}{8}} $


*$1+2+3+...+k+(k+1)<\color{red}{\frac{(2k+1)²}{8}+(k+1)} $
Now, following the way how Apostol proceeded from this point in other proofs like this one in the book, I have to prove:
$\frac{(2k+1)²}{8}+(k+1)<\frac{(2(k+1)+1)²}{8}$
But

*

*$\frac{(2k+1)²}{8}+(k+1)=\frac{4k²+12k+9}{8}$


*$\frac{(2(k+1)+1)²}{8}=\frac{4k²+12k+9}{8}$
So $\color{blue}{\frac{(2k+1)²}{8}+(k+1)<\frac{(2(k+1)+1)²}{8}}$ is false, but according to this other proof in the book


$\color{blue}{\frac{(2k+1)²}{8}+(k+1)<\frac{(2(k+1)+1)²}{8}}$ should be true, for the inequality to hold, but the result I'm getting tells me

$$1+2+3+...+n=\frac{(2n+1)²}{8} $$

which seems to be false according to

$A(1) = 1 < \frac{9}{8}$
$A(2) = 1+2=3 < \frac{25}{8}$

and

So, what am I doing wrong?
 [/INST] You are almost done. Note that you should show that
$$1+2+3+\dots+k+(k+1)<\frac{(2k+1)^2}{8}+(k+1)\color{red}{\leq }\frac{(2(k+1)+1)^2}{8}.$$
which implies that
$$1+2+3+\dots+k+(k+1)< \frac{(2(k+1)+1)^2}{8}.$$
So showing that
$$\frac{(2k+1)^2}{8}+(k+1)=\frac{(2(k+1)+1)^2}{8}$$
concludes the proof.
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