General ⇒ Runaway argument error

[ThomasAngel]

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$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{k^2} \leq 2 - \frac{1}{k} .$



$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{k^2} + \frac{1}{(k+1)^2}\leq 2 - \frac{1}{k} + \frac{1}{(k+1)^2} $

$\sum_{i=1}^k+1 \frac{1}{i^[2]} \leq 2 - \frac{1}{k} + \frac{1}{(k+1)^2} $

\[
\sum_{i=1}^k+1 \frac{1}{i^[2]} \leq 2 - \frac { (k+1)^{2}}{(k)(k+1)^{2}} + \frac{k}{k(k+1)^{2}} 
\]


$= 2 - \frac {(k+ k^{2} +2k+1)}{(k)(k+1)^{2}}$

$=2 - \frac{ (k^{2} +k)}{(k){(k+1)^{2}} - \frac{2k+1}{(k)(k+1)^{2}}$

$=2 - \frac{ (k)(k+1))}{(k){(k+1)^{2}} - \frac{2k+1}{(k)(k+1)^{2}}$

$2- \frac{1}{k+1} - \frac{2k+1}{{(k)(k+1)^[2]}$

In order for P(k) to imply P(k+1), 

$\sum_{i=1}^k+1 \frac{1}{i^[2]} \leq 2 - \frac{1}{k+1}$

$2 - \frac{1}{k+1} = 2 - \frac{k(k+1)}{(k)(k+1)^[2]}$

$2- \frac{1}{k+1} - \frac{2k+1}{{(k)(k+1)^[2]}  \leq 2 - \frac{1}{k+1}$

[Bartman]

In several denominators one opening brace is too much.

The first occurance:

error-prone

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$= 2 - \frac {(k+ k^{2} +2k+1)}{{(k)(k+1)^2}$

error-free

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$= 2 - \frac {(k+ k^{2} +2k+1)}{(k)(k+1)^2}$

Also, for a one-digit number in the exponent, you don't have to use parentheses, but if you do, use a couple of braces instead of brackets.