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Butoxors [25]
2 years ago
6

20%20%5Cdisplaystyle%20%5Crm%20%5Csum_%7B%20%20n_%7B2%7D%20%3D%201%7D%5E%20%5Cinfty%20%20%20%5Cdots%20%5Cdisplaystyle%20%5Crm%20%5Csum_%7B%20%20n_%7B2022%7D%20%3D%201%7D%5E%20%5Cinfty%20%20%20%5Cfrac%7B1%7D%7Bn_%7B1%7D%20n_2%20%5Cdots%20n_%7B2022%7D%28n_%7B1%7D%20%2B%20n_2%20%20%2B%20%5Cdots%20%2B%20%20n_%7B2022%7D%29%7D%20" id="TexFormula1" title=" \displaystyle \rm \sum_{ n_{ 1} = 1}^ \infty \displaystyle \rm \sum_{ n_{2} = 1}^ \infty \dots \displaystyle \rm \sum_{ n_{2022} = 1}^ \infty \frac{1}{n_{1} n_2 \dots n_{2022}(n_{1} + n_2 + \dots + n_{2022})} " alt=" \displaystyle \rm \sum_{ n_{ 1} = 1}^ \infty \displaystyle \rm \sum_{ n_{2} = 1}^ \infty \dots \displaystyle \rm \sum_{ n_{2022} = 1}^ \infty \frac{1}{n_{1} n_2 \dots n_{2022}(n_{1} + n_2 + \dots + n_{2022})} " align="absmiddle" class="latex-formula">​
Mathematics
1 answer:
Kay [80]2 years ago
8 0

As a simpler example, consider the iterated sum with only 2 indices,

\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac1{n_1n_2(n_1+n_2)}

(The case with just one index is pretty simple, as it reduces to ζ(2) = π²/6.)

Let

\displaystyle f(x) = \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac{x^{n_1+n_2}}{n_1n_2(n_1+n_2)}

Differentiating and multiplying by x, we get

\displaystyle x f'(x) = \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac{x^{n_1+n_2}}{n_1n_2} \\\\ = \left(\sum_{n_1=1}^\infty\frac{x^{n_1}}{n_1}\right) \left(\sum_{n_2=1}^\infty \frac{x^{n_2}}{n_2}\right) \\\\ = (-\ln(1-x))^2 = \ln^2(1-x)

\implies f'(x) = \dfrac{\ln^2(1-x)}x

By the fundamental theorem of calculus (observing that letting x = 0 in the sum makes it vanish), we have

f(x) = \displaystyle \int_0^x \frac{\ln^2(1-t)}t \, dt

If we let x approach 1 from below, f(x) will converge to the double sum and

\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac1{n_1n_2(n_1+n_2)} = \int_0^1 \frac{\ln^2(1-x)}x \, dx

In the integral, substitute x\mapsto1-x, use the power series expansion for 1/(1 - x), and integrate by parts twice.

\displaystyle \int_0^1 \frac{\ln^2(1-x)}x \, dx = \int_0^1 \frac{\ln^2(x)}{1-x} \, dx \\\\ = \sum_{m=0}^\infty \int_0^1 x^m \ln^2(x) \, dx \\\\ = \sum_{m=0}^\infty -\frac2{m+1} \int_0^1 x^m \ln(x) \, dx \\\\ = \sum_{m=0}^\infty \frac2{(m+1)^2} \int_0^1 x^m \, dx \\\\ = 2 \sum_{m=0}^\infty \frac1{(m+1)^3} \\\\ = 2 \sum_{m=1}^\infty \frac1{m^3} = 2\zeta(3)

We can generalize this method to k indices to show that

\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty \frac1{n_1n_2\cdots n_k(n_1+n_2+\cdots+n_k)} = (-1)^k \int_0^1 \frac{\ln^k(1-x)}x \, dx \\\\ = k!\,\zeta(k+1) = \Gamma(k+1)\zeta(k+1)

Then the sum we want is

\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_{2022}=1}^\infty \frac1{n_1n_2\cdots n_{2022}(n_1+n_2+\cdots+n_{2022})} = \boxed{\Gamma(2023)\zeta(2023)}

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The perimeter of a rectangle is 140 feet. The length of the rectangle is 6 more than 2 times the width. What is the length of th
Alja [10]
The equations we get are
l = 6 + 2w (We get this from "The length of the rectangle is 6 more than 2 times the width.")
and
2l + 2w = 140 (We get this from the perimeter. Two times the length plus two times the width equals the perimeter of a quadrilateral.)

The first equation can be written as
l - 2w = 6

So now we have
2l + 2w = 140
l - 2w = 6
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Add the two equations.
We get 3l = 146

Divide by 3 on both sides.
l = 48\frac{2}{3}

Your answer is 48\frac{2}{3}.
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3 years ago
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puteri [66]
1st. is definitely 150.
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3 years ago
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WILL MARK BRAINLIEST!!!!!!!!! <br> what is 5/1 as a whole number
MariettaO [177]

Answer:

5

Step-by-step explanation:

5÷1

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3 0
3 years ago
How do i solve z^2+9=10z
Musya8 [376]

Answer:

z = 1, z = 9

Step-by-step explanation:

Given

z² + 9 = 10z ( subtract 10z from both sides )

z² - 10z + 9 = 0

Consider the factors of the constant term (+ 9) which sum to give the coefficient of the z- term (- 10)

The factors are - 1 and - 9, since

- 1 × - 9 = 9 and - 1 - 9 = - 10, thus

(z - 1)(z - 9) = 0 ← in factored form

Equate each factor to zero and solve for z

z - 1 = 0 ⇒ z = 1

z - 9 = 0 ⇒ z = 9

3 0
3 years ago
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