Exploring the Convergence and Divergence of Series in Calculus

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In calculus, understanding the behavior of series, specifically whether they converge or diverge, is crucial. This article delves into the analysis of the series (Ssum_{n1}^{infty}frac{1}{(22n-1)^2}frac{1}{8n^3}), providing a comprehensive explanation using key tests and principles.

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Convergence of given series

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We first examine whether the series

(Ssum_{n1}^{infty}frac{1}{(22n-1)^2}frac{1}{8n^3}) converges. To do this, we can break it down into two sub-series:

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(sum_{n1}^{infty}frac{1}{(22n-1)^2}) and (sum_{n1}^{infty}frac{1}{8n^3}).

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Convergence of (sum_{n1}^{infty}frac{1}{(22n-1)^2})

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The series (sum_{n1}^{infty}frac{1}{n^2}) is a well-known convergent p-series with (p2). Since the terms of (sum_{n1}^{infty}frac{1}{(22n-1)^2}) are bounded above by the terms of (sum_{n1}^{infty}frac{1}{n^2}), the comparison test confirms the convergence of (sum_{n1}^{infty}frac{1}{(22n-1)^2}).

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Additionally, we consider (sum_{n1}^{infty}frac{1}{(2n-1)^2}) as a sub-series of (sum_{n1}^{infty}frac{1}{n^2}). Since both series are convergent, the original series (sum_{n1}^{infty}frac{1}{(22n-1)^2}) must also converge.

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Convergence of (sum_{n1}^{infty}frac{1}{8n^3})

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The series (sum_{n1}^{infty}frac{1}{n^3}) is also a p-series with (p3), which converges. The factor of 8 in the denominator is simply a constant multiplier, so it does not affect the convergence of the series.

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Thus, the original series (S) can be rewritten as:

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(Sfrac{1}{2}sum_{n1}^{infty}frac{1}{(22n-1)^2}frac{1}{8}sum_{n1}^{infty}frac{1}{n^3})

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Since this is the sum of two convergent series, it is also convergent.

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Euler's Proof of Divergence of Zeta Series at 1

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Leonhard Euler was the first to recognize that the series of reciprocals of primes, (sum_{p le x}frac{1}{p}), diverges.

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Euler's groundbreaking proof relies on the study of the product of 1 over (1 - 1/p) for all primes p not greater than x, denoted by

(S_n prod_{i1}^nleft(1-frac{1}{p_i}right)^{-1}), where (p_n) is the largest prime not greater than x.

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Theorem

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The theorem states that the sum of the reciprocals of all primes satisfies the inequality

(left(sum_{p le x}frac{1}{p}right) - loglog x - 1)

The result is bounded by a constant independent of x.

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Proof

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To prove this, Euler starts by noting the inequality

(log x ge x)

for all x 1. From this, it follows that for any prime p:

(logleft(1 - frac{1}{p}right) le -frac{1}{p-1})

Multiplying both sides by -1, we get

(1 - frac{1}{p} le -logleft(1 - frac{1}{p}right)^{-1} le frac{1}{p-1})

Summing from i 1 to n, the inequality simplifies to

(0 le log S_n - sum_{i1}^n frac{1}{p_i} le sum_{i1}^n frac{1}{p_i(p_i - 1)} le sum_{i2}^{lfloor xrfloor} frac{1}{ii - 1} 1 - frac{1}{lfloor xrfloor})

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This completes the proof by showing that as x approaches infinity, the sum of reciprocals of primes diverges, closely related to the natural logarithm of x.