Sum to the infinity of a series

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SUMMARY

The forum discussion focuses on calculating the infinite series \(\sum_{n=1}^{\infty} \frac{1}{1+(n-1)^2} \left(\frac{1}{3}\right)^{2+(n-1)^2}\). Participants suggest transforming the series by starting from \(n=0\) to simplify the expression. The use of arithmetico-geometric series and the logarithmic expansion of \(\ln(1+y)\) is emphasized as a method to approach the solution. Additionally, the integral \(\int_0^x t^{2n} dt\) is mentioned as a relevant tool for evaluating the series.

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  • Understanding of infinite series and convergence
  • Familiarity with arithmetico-geometric series
  • Knowledge of logarithmic functions and their expansions
  • Basic calculus, specifically integration techniques
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  • Study the properties of arithmetico-geometric series
  • Learn about the expansion of \(\ln(1+y)\) and its applications
  • Explore techniques for evaluating infinite series
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justwild
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Homework Statement



to find the value of [itex]\sum[/itex] over n=1 to ∞ of [1/{1+(n-1)2}](1/3)[itex]^{2+(n-1)2}[/itex]

Homework Equations





The Attempt at a Solution


I have tried to solve in the way the arithmetico-geometric series are solved and tried to bring it in the form of the expansion of ln(1+y), because the answer has logarithmic term in it.
 
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justwild said:

Homework Statement



to find the value of [itex]\sum[/itex] over n=1 to ∞ of [1/{1+(n-1)2}](1/3)[itex]^{2+(n-1)2}[/itex]

Homework Equations


The Attempt at a Solution


I have tried to solve in the way the arithmetico-geometric series are solved and tried to bring it in the form of the expansion of ln(1+y), because the answer has logarithmic term in it.

Hints:

Get rid of the (n-1) by starting the sum off from n = 0, that'll make things clearer.

Observe that [itex]{(\frac{1}{3})}^{2 + 2n} =\frac{1}{3}.{(\frac{1}{3})}^{2n + 1}[/itex]

What's [itex]\int_0^x t^{2n} dt[/itex]?
 

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