Σn/[(n+1)(n+2)(n+3)] Series

Thread starter Feynmanfan
Start date Apr 7, 2004
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Series

In summary, the formula for calculating the sum of the Σn/[(n+1)(n+2)(n+3)] series is S = (3n+2)/[4(n+1)(n+2)]. The series is a telescoping series and converges to a finite value. The value of n does not affect the convergence. The series can be written in a closed form and is related to the Σ1/n^3 series.

Apr 7, 2004

Feynmanfan

Could anybody help me with this

Σn/[(n+1)(n+2)(n+3)]

Thank you guys!

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Apr 8, 2004

arcnets

I guess the first step is to rewrite this into
A/(n+1) + B/(n+2) + C/(n+3)
where A, B, C are numbers.

Apr 8, 2004

cookiemonster

https://www.physicsforums.com/showthread.php?t=18630

...?

cookiemonster

1. What is the formula for calculating the sum of the Σn/[(n+1)(n+2)(n+3)] series?

The formula for calculating the sum of the Σn/[(n+1)(n+2)(n+3)] series is S = (3n+2)/[4(n+1)(n+2)].

2. What is the convergence of the Σn/[(n+1)(n+2)(n+3)] series?

The Σn/[(n+1)(n+2)(n+3)] series is a telescoping series, meaning that most of the terms in the series cancel out. Therefore, the series converges to a finite value.

3. How does the value of n affect the convergence of the Σn/[(n+1)(n+2)(n+3)] series?

The value of n does not affect the convergence of the Σn/[(n+1)(n+2)(n+3)] series, as long as it is a positive integer. The series will still converge to a finite value.

4. Can the Σn/[(n+1)(n+2)(n+3)] series be written in a closed form?

Yes, the Σn/[(n+1)(n+2)(n+3)] series can be written in a closed form as S = (3n+2)/[4(n+1)(n+2)].

5. How is the Σn/[(n+1)(n+2)(n+3)] series related to other series?

The Σn/[(n+1)(n+2)(n+3)] series is related to the Σ1/n^3 series, as both are examples of telescoping series. However, the Σn/[(n+1)(n+2)(n+3)] series converges to a finite value, while the Σ1/n^3 series diverges.