Undergrad Is Wolfram Alpha Wrong About This Infinite Sum?

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SUMMARY

The discussion centers around the infinite sum $$\sum_{n=2}^{\infty }\frac{3^n-1}{4^n}\zeta (n+1)$$ and its evaluation using Wolfram Alpha, which approximates the sum to 2.319125. Participants conclude that the book referenced likely contains a typographical error, as adjusting the sum to start at 1 yields a result converging to π (approximately 3.14159265). The consensus is that the original formulation in the book is incorrect.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with the Riemann zeta function, specifically $\zeta(n)$
  • Basic knowledge of mathematical notation and summation
  • Experience using computational tools like Wolfram Alpha
NEXT STEPS
  • Investigate the properties of the Riemann zeta function, particularly $\zeta(n+1)$
  • Learn about convergence criteria for infinite series
  • Explore how to use Wolfram Alpha for evaluating complex mathematical expressions
  • Study common typographical errors in mathematical literature and their implications
USEFUL FOR

Mathematicians, students studying advanced calculus, and anyone interested in the evaluation of infinite series and the use of computational tools in mathematics.

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Hi, I'm reading a book of math and in one page says:
$$\sum_{n=2}^{\infty }\frac{3^n-1}{4^n}\zeta (n+1)=\pi$$
I tried to solve this,as I could not solved I requested to wolfram alpha and told me that this sum is
approximately equal to 2.319125.
https://www.wolframalpha.com/input/?i=sum+2+to+infinity+(3^k-1)/4^k+zeta(k+1)
so i do not know if wolfram alpha is wrong or the book have a mistake and I like to know which is wrong.i tried some other values and the most proximate to pi is minus gamma, but I not quite shure and wolfram alpha calculation time expired u.u.

so i do not know if this is true
$$\sum_{n=2}^{\infty }\frac{3^n-1}{4^n}\zeta (n-\gamma )\approx \pi $$
 
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