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Conclude if the given series converges or diverges and justify the conclusion

Test the following series for convergence or divergence. Justify the decision.

    \[ \sum_{n=1}^{\infty} \frac{\sqrt{2n-1} \log(4n+1)}{n(n+1)}. \]


    \[ a_n = \frac{\sqrt{2n-1} \log (4n+1)}{n(n+1)}, \qquad b_n = \frac{n^{\varepsilon}}{n^{\frac{3}{2}}} \quad 0 < \varepsilon < \frac{1}{2}. \]

Then, the series \sum b_n converges by Example #1 on page 398 of Apostol,

    \[ b_n = \frac{1}{n^{\frac{3}{2} - \varepsilon}} \quad \implies \quad b_n = \frac{1}{n^s} \]

where s > 1 (since \varepsilon < \frac{1}{2}). Then consider the limit,

    \begin{align*}  \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\frac{\sqrt{2n-1} \log(4n+1)}{n(n+1)}}{\frac{n^{\varepsilon}}{n^{\frac{3}{2}}}} \\[9pt]  &= \lim_{n \to \infty} \frac{n^{\frac{3}{2}} \sqrt{2n-1} \log (4n+1)}{n^{\varepsilon+1} (n+1)} \\[9pt]  &= \lim_{n \to \infty} \left( \frac{\sqrt{n} \sqrt{2n-1}}{n+1} \cdot \frac{\log(4n+1)}{n^{\varepsilon}} \right). \end{align*}

The limits of each of the terms in the product exist (as we show below) so the limit of the product is the product of the limits,

    \begin{align*}  &= \left( \lim_{n \to \infty} \frac{\sqrt{n}\sqrt{2n-1}}{n+1} \right) \left( \lim_{n \to \infty} \frac{\log(4n+1)}{n^{\varepsilon}} \right) \\[9pt]  &= \left( \lim_{n \to \infty} \frac{\sqrt{2n^2 - n}}{n+1} \right) \left( \lim_{n \to \infty} \frac{\log(4n+1)}{n^{\varepsilon}} \right) \\[9pt]  &= \left( \lim_{n \to \infty} \frac{\sqrt{2 - \frac{1}{n}}}{1 + \frac{1}{n}} \right) \left( \lim_{n \to \infty} \frac{\log(4n+1)}{n^{\varepsilon}} \\[9pt]  &= \sqrt{2} \cdot 0 \\[9pt]  &= 0. \end{align*}

The limit \lim_{n \to \infty} \frac{\log (4n+1)}{n^{\varepsilon}} = 0 since we know

    \[ \lim_{n \to \infty} \frac{(\log n)^a}{n^b} = 0 \]

for all a>0, b>0. Therefore, we have

    \[ \lim_{n \to \infty} \frac{a_n}{b_n} = 0. \]

By Theorem 10.9 (see the note after the proof of the theorem on page 396 of Apostol), we then have the convergence of \sum b_n implies the convergence of \sum a_n. Hence,

    \[ \sum_{n=0}^{\infty} \frac{\sqrt{2n-1} \log(4n+1)}{n(n+1)} \]



  1. MathlessRick says:

    Well as n -> infinity
    Sqrt(2n-1) ~ sqrt(2n)
    n(n+1) ~ n^2
    log(4n+1) ~ log(4n) = log(4) + log(n)

    So just solve for
    ln4 * Sqrt(2n)/n^2 + sqrt(2)*ln(n)/n^(3/2)

    Which clearly converges for both parts -> 0

    So the sum must converge by theorem 10.10… right?

    • Mohammad Azad says:

      Funny how Apostol clearly warned the reader from falling in this fallacy, if the terms of a series tend to zero, that doesn’t mean the series converges. The harmonic series is the classic example.

  2. Artem says:

    This can be done also simpler (from my point of view): notice that the logarithm term is less than n^{1/8} for all n > some N. Then, sqrt{2n – 1} < sqrt{2n}. From this, we find the identity a_n < \sqrt{2}b_n, which leads to Riemann zeta with power 11/8. Thus, by comparison test the series is convergent.

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