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$$ \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{\sqrt{4n+1}} - \frac{1}{\sqrt{5n+1}}\right) $$

Will someone please help me validate my way?

After moving to a common denominator, and multiplying by the "conjugate", we obtain the series: $$ \sum_{n=1}^{\infty} (-1)^n \left( \frac{n}{ \sqrt{5n+1} \sqrt{4n+1} (\sqrt{4n+1}+\sqrt{5n+1}) } \right) $$

as for the absolute convergence: we can use limit comparison test with $\frac{1}{n^{0.5}}$ to obtain the series diverges. As for convergence- this is a Leibnitz series, so it converges.

Am I right?

Thanks

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    $\begingroup$ It is clear that the terms go to $0$ in absolute value. You have not verified explicily that the terms decrease in absolute value. All the rest is fine. $\endgroup$ – André Nicolas Mar 30 '16 at 19:28
  • $\begingroup$ Yes it does absolutely diverge. For the convergence you just have to show that $\frac{1}{\sqrt{4n+1}}-\frac{1}{\sqrt{5n+1}}$ tends to zero (obvious) and that it is definitively decreasing for $n\rightarrow +\infty$. Have you done that? $\endgroup$ – Nicolò Mar 30 '16 at 19:29
  • $\begingroup$ I can show it is decreasing by moving to a function and differentiating, right? $\endgroup$ – yehushua Mar 30 '16 at 20:08
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$$ \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^n } \frac{1} {{\sqrt {4n + 1} }} $$ is convergent by Leibniz. The same is true for $$ \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^n } \frac{1} {{\sqrt {5n + 1} }} $$ thus the given series is convergent because is the difference of two convergent series.

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