Test the following series for convergence or divergence. Justify the decision.
![\[ \sum_{n=1}^{\infty} \frac{\sqrt{2n-1} \log(4n+1)}{n(n+1)}. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-dbcd3f95871d70b2483f833fa808876b_l3.png "Rendered by QuickLaTeX.com")
Let
![\[ a_n = \frac{\sqrt{2n-1} \log (4n+1)}{n(n+1)}, \qquad b_n = \frac{n^{\varepsilon}}{n^{\frac{3}{2}}} \quad 0 < \varepsilon < \frac{1}{2}. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8fa62a323ee852ce2b6971241b247e7d_l3.png "Rendered by QuickLaTeX.com")
Then, the series 
converges by Example #1 on page 398 of Apostol,
![\[ b_n = \frac{1}{n^{\frac{3}{2} - \varepsilon}} \quad \implies \quad b_n = \frac{1}{n^s} \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-af70d843e68d1abc6a75c1178c433697_l3.png "Rendered by QuickLaTeX.com")
where 
(since 
). 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*}](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-93b7a4489f0c9f36f189c93ca5a94ce4_l3.png "Rendered by QuickLaTeX.com")
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*}](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-979c78458e932da1fcf7a80344e495fa_l3.png "Rendered by QuickLaTeX.com")
The limit 
since we know
![\[ \lim_{n \to \infty} \frac{(\log n)^a}{n^b} = 0 \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a4164711e47936e6be1c998959141e8e_l3.png "Rendered by QuickLaTeX.com")
for all 
, 
. Therefore, we have
![\[ \lim_{n \to \infty} \frac{a_n}{b_n} = 0. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7c348ebe954fc6b79c24cedf941f9371_l3.png "Rendered by QuickLaTeX.com")
By Theorem 10.9 (see the note after the proof of the theorem on page 396 of Apostol), we then have the convergence of implies the convergence of 
. Hence,
![\[ \sum_{n=0}^{\infty} \frac{\sqrt{2n-1} \log(4n+1)}{n(n+1)} \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-07349fb6b245fafbb2eb63edadff9bfd_l3.png "Rendered by QuickLaTeX.com")
converges.
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.