Approximating constant π² to within error
You don't need to compare prev and new during each iteration.

The difference between the new and the previous sum is simply the current term : $$\frac{8}{(2i+1)^2}$$

If you want this term to be smaller than error, you can solve:

$$\mathrm{error} &gt; \frac{8}{(2i+1)^2}\\
\iff (2i+1)^2 &gt; \frac{8}{error}\\
\iff 2i+1 &gt; \sqrt{\frac{8}{error}}\\
\iff i &gt; \frac{\sqrt{\frac{8}{error}} - 1}{2}\\
$$

Now that you know how many terms your series should have, you can return the result directly:

def approx_pi_squared(error):
    n = int(((8 / error)**0.5 - 1) / 2) + 1
    return sum(8 / (2 * i + 1)**2 for i in range(n))


Better formulas

Adding a delta

Note that error represents how small the terms are, not how close approx_pi_squared is from π²:

&gt;&gt;&gt; import math
&gt;&gt;&gt; approx_pi_squared(1e-10)
9.869590258918535
&gt;&gt;&gt; math.pi**2 - approx_pi_squared(1e-7)
0.0004472271895057389
&gt;&gt;&gt; math.pi**2 - approx_pi_squared(1e-10)
1.414217082285063e-05


Even with more than 140 000 terms, the series only gives the 3 first digits of π². This formula is very simple but converges too slowly.

What's very interesting, though, is that the difference between math.pi**2 and approx_pi_squared(error) seems very close to \$\sqrt{2\mathrm{error}}\$. It seems to hold true for any error, so we could update the function:

def approx_pi_squared(error):
    n = int(((8 / error)**0.5 - 1) / 2) + 1
    delta = (2 * error)**0.5
    return sum(8 / (2 * i + 1)**2 for i in range(n)) + delta


approx_pi_squared(1e-10) now returns 10 correct digits for π².

This new formula is unproven, so use at your own risk!

BBP-Type Formula

There are many π² formulas, so feel free to pick another one. For example:

def approx_pi_squared(error):
    n = int(((12 / error)**0.5 - 1) / 2) + 1
    return 12 * sum((-1)**i / (i + 1)**2 for i in range(n))


error seems to have the same order of magnitude as math.pi**2 - approx_pi_squared(error) now:

&gt;&gt;&gt; math.pi**2 - approx_pi_squared(1e-9)
2.0001476030984122e-09
&gt;&gt;&gt; math.pi**2 - approx_pi_squared(1e-10)
-1.9977974829998857e-10


delta looks like (-1)**n * 2 * error.

With Sympy

You can delegate the job to sympy and be sure you'll get a correct result with arbitrary precision:

&gt;&gt;&gt; from sympy import Sum, N, pi, oo, init_printing
&gt;&gt;&gt; from sympy.abc import i, n
&gt;&gt;&gt; init_printing()
&gt;&gt;&gt; Sum(8/(2*i+1)**2, (i, 0, oo))
  ∞             
 ____           
 ╲              
  ╲       8     
   ╲  ──────────
   ╱           2
  ╱   (2⋅i + 1) 
 ╱              
 ‾‾‾‾           
i = 0  
&gt;&gt;&gt; N(Sum(8/(2*i+1)**2, (i, 0, oo)), 100)
9.869604401089358618834490999876151135313699407240790626413349376220044822419205243001773403718552232
&gt;&gt;&gt; N(pi**2, 100)
9.869604401089358618834490999876151135313699407240790626413349376220044822419205243001773403718552232