Post your math problems here #3 [And solutions to #2]

@mapofdragons said in #13:

Problem: if 11987 + 21986 + 3*1985 + ... + 1986 * 2 + 1987 * 1 = 1987 * 994 * x, then what is x?

x = 663

@tpr said in #20:

"I suppose"

• I put in 2 uppercase 2, but this site renders it as plain 2
  I edited above.
  The series is divergent.
  You have computed the sum of the series sum_{n=1}^{infinity} (n-1)/n^{2} and this series coincides
  with the series written in #14 if we understand "^" in #14 as multiplication. But if we understand "^"
  in #14 as exponentiation, then the series written in #14 is convergent.

@E4_Entangler said in #14:

What is the sum of the infinite series 1/(2^2) + 2/(3^3) + 3/(4^4) +4/(5^5) + 5/(6^6)...? I don't know the answer at all lol

This is convergent, no doubt about it, since the finite sum until n is increasing and bounded above by

1/2! + 2/3! + ... + n/(n+1)!

Since n^n > n! ( regarding the discussion above, I've never, ever, read smth where ^ means multiplication. Of course it's exponentiation).

Noting that n/(n+1) < 1, the above is in turn bounded above by

1/1! + 1/2! + ... + 1/n!

Which converges to e. Therefore that series s converges and is bounded by

1/4 < s <= e

This should test your knowledge of calculus. And I do know the solution and how to solve for it. What is the i-th root of -1? Ofc this can be written (-1)^(1/i).

"read smth where ^ means multiplication"

• It must be my eyesight... I solved another question than the one that was asked. My bad.

"What is the i-th root of -1? Ofc this can be written (-1)^(1/i)."

• (-1)^(1/i) = exp (ln(-1) / i) = exp (ln (exp(i pi))/i) = exp (pi)

@m011235 said in #23:

This is convergent, no doubt about it, since the finite sum until n is increasing and bounded above by

1/2! + 2/3! + ... + n/(n+1)!

Since n^n > n! ( regarding the discussion above, I've never, ever, read smth where ^ means multiplication. Of course it's exponentiation).

Noting that n/(n+1) < 1, the above is in turn bounded above by

1/1! + 1/2! + ... + 1/n!

Which converges to e. Therefore that series s converges and is bounded by

1/4 < s <= e
Just one thing: the series 1 + 1/1! + 1/2! + 1/3!... converges to e, not 1/1! + 1/2!... . Therefore the bounds would be 1/4 < s < e-2.
Also, checking a calculator, this series seems to approach roughly 0.33718, but I want to see if there exists a mathematical way to solve this.

@E4_Entangler said in #27:

Just one thing: the series 1 + 1/1! + 1/2! + 1/3!... converges to e, not 1/1! + 1/2!... .

Good point.

Therefore the bounds would be 1/4 < s < e-2.

That would be 1/4 < s < e-1, at least with my bound, maybe you found the better e-2?

Also, checking a calculator, this series seems to approach roughly 0.33718, but I want to see if there exists a mathematical way to solve this.

Yeah, that series converges lightning fast.

@m011235 the series which converges to e is 1 + 1/1! + 1/2! + 1/3!... and since the series is 1/2^2 + 2/3^3..., this series is bounded by 1/2! + 1/3! + 1/4!... which is e-2.

@E4_Entangler yes this is true. This can be further improved by noting that (k-1)^2 > 0 for k>=2, which leads to (k+1)^2 > 4k then (k+1)^k > 2kk! and finally k/(k+1)^(k+1) < 1/2 * 1/(k+1)!.

Thus the series is bounded above by 1/2 * ( 1/2! + 1/3! + ...) which is (e-2)/2, so approximating

0.25 < s < 0.35 (I'm using e=2.7 lol)