Thread 16751904

10 posts 8 images /sci/

Anonymous 8/15/2025, 5:36:03 PM No.16751904 >>16752301 >>16752357

[math]
\sum_{n=1}^{\infty} \frac{10^{2^{n-1}} - 1}{9 \cdot 10^{2^{n-1} + n - 2}} = ?
[/math]

Anonymous 8/16/2025, 12:12:16 AM No.16752301

>>16751904 (OP)
UH... 7?

Anonymous 8/16/2025, 1:20:11 AM No.16752357

>>16751904 (OP)
0 because 10^(2^(n-1)) is a factor of 10^(2^((n-1))+(n-2)) and therefore the numerator simplifies to 1-1 which is 0 for all n.

Anonymous 8/16/2025, 1:51:20 AM No.16752385 >>16752434

quick maths.png md5: 63b4554f... 🔍

Anonymous 8/16/2025, 3:00:20 AM No.16752434 >>16752438 >>16753147

>>16752385
calculating the sum of the first 20 terms with python gets me 1.1223445678901234 so this seems inaccurate

Anonymous 8/16/2025, 3:08:18 AM No.16752438

>>16752434
nvm I'm off but the image is off as well.

Anonymous 8/16/2025, 11:31:07 PM No.16753147 >>16754852

to n=12.gif md5: 3e0d7f74... 🔍

>>16752434
https://www.wolframalpha.com/input?i=Sum%5B%2810%5E2%5E%28n+-+1%29+-+1%29%2F%289+10%5E%282%5E%28n+-+1%29+%2B+n+-+2%29%29%2C+%7Bn%2C+1%2C+12%7D%5D

Anonymous 8/18/2025, 9:57:12 PM No.16754835

Bump

Anonymous 8/18/2025, 10:16:07 PM No.16754852 >>16754854

Screenshot_2025-08-18_22.15.17.png md5: fae98128... 🔍

>>16753147
That is rather curious.

Anonymous 8/18/2025, 10:17:38 PM No.16754854

>>16754852
Minor mistake, that last 8 in the first row of the n = 12 section should be one row lower to actually give us nine 8s.
Anyway I don't really understand infinite sums so maybe this is only curious to me.