What is the Convergent Series Sum Formula?

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SUMMARY

The infinite sum of the series S_n = ∑_{j=1}^∞ β^j j converges to the formula S_n = β / (1 - β)^2, where β is a constant with |β| < 1. This result is derived using the properties of arithmetico-geometric series. The discussion confirms the validity of the formula through a step-by-step manipulation of the series and its terms, leading to the final expression. The solution is established as correct and provides a clear method for calculating the sum of similar series.

PREREQUISITES
  • Understanding of infinite series and convergence criteria
  • Familiarity with arithmetico-geometric series
  • Basic algebraic manipulation skills
  • Knowledge of the geometric series formula
NEXT STEPS
  • Study the properties of arithmetico-geometric series in detail
  • Learn about convergence tests for infinite series
  • Explore applications of the sum formula in calculus
  • Investigate related series and their sums, such as power series
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in series convergence and summation techniques.

jediwhelan
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Homework Statement



Dear All,

I have a series that I know to converge but for which I can't work out the infinite sum. It should be something simple.

[tex] S_n = \sum_{j=1}^\infty \beta^j j[/tex]

Can somebody help me with this?

I think the solution is:

[tex] \frac{\beta}{(1-\beta)^2}[/tex]
 
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It looks right:

[tex]S_n = \beta + 2\beta^2 + 3\beta^3 + \cdots[/tex]

[tex]\frac{S_n}{\beta} = 1 + 2\beta + 3\beta^2 + \cdots[/tex]

[tex](\frac{1}{\beta}-1) S_n = 1 + \beta + \beta^2 + \cdots[/tex]

[tex](\frac{1}{\beta}-1) S_n = \frac{1}{1-\beta}[/tex]

[tex](\frac{1-\beta}{\beta}) S_n = \frac{1}{1-\beta}[/tex]

[tex]S_n = \frac{\beta}{(1-\beta)^2}[/tex]

It's called an arithmetico-geometric series I think,

[tex]\displaystyle\sum_{n=0}^{\infty}(a+nd)r^n = \frac{a}{1-r} + \frac{rd}{(1-r)^2}[/tex]
 
brilliant. I was trying something like that but couldn't get it.

Thanks for the quick reply.

Paul
 

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