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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.

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


The series converges. To show this, first we note that

    \[ \frac{\log n}{n \sqrt{n+1}} \leq \frac{\log n}{n^{\frac{3}{2}}}. \]

Thus, if we can show that the series

    \[ \sum_{n=2}^{\infty} \frac{\log n}{n^{\frac{3}{2}}} \]

converges then the series in question converges by the comparison test. To show that this series converges we apply the limit comparison test with

    \[ a_n = \frac{\log n}{n^{\frac{3}{2}}}, \qquad b_n = \frac{n^{\varepsilon}}{n^{\frac{3}{2}}} \qquad 0 < \varepsilon < \frac{1}{2}. \]

We know that \sum b_n converges since it is of the form \sum \frac{1}{n^s} for s > 1. Furthermore,

    \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\log n}{n^{\varepsilon}} = 0 \]

for \varepsilon > 0. Thus,

    \[ \sum_{n=2}^{\infty} \frac{\log n}{n^{\frac{3}{2}}} \]

converges which implies the convergence of

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

2 comments

  1. Anonymous says:

    Si no estoy mal y segun el apostol, el criterio de comparacion a limite funciona cuando el limite da 1, en este caso el limite nos dio 0 :(

    • Anonymous says:

      Si, cuando da 1 sucede que la serie de arriba y la de abajo se comportan igual
      O sea, si una diverge la otra también o si una converge la otra también
      Si el límite da 0 indica que si la de abajo converge la de arriba también (si la de abajo diverge no indica nada)

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