Extremely simple question regarding the convergence of a series

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    Convergence Series
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SUMMARY

The discussion centers on the convergence of the alternating series defined by the formula \(\sum\limits_{j=1}^\infty(-1)^j\frac{\sqrt{j}}{j+5}\). The limit of the terms approaches zero, satisfying the first condition of the Alternating Series Test. However, the challenge lies in proving that the non-alternating series converges, specifically whether each term is less than the previous term. The user is advised to utilize the relationship between squared terms and apply the comparison test for rigorous analysis.

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TheFerruccio
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First off, this is not a homework question. I am helping someone with an alternating series, and, for some reason, I am finding myself completely baffled by this one.

I have an alternating series that takes the form:

[itex]\sum\limits_{j=1}^\infty(-1)^j\frac{\sqrt{j}}{j+5}[/itex]

I know that the limit approaches zero, so the first requirement of the alternating series test is filled.

However, I have been doing all sorts of algebra, whether it's fractions, differences, comparing two fractions, in order to figure out whether the non alternating series converges always, meaning, will each subsequent term always be less than the previous term.

I have not been able to figure this part out. I keep ending up with funky comparisons that are not mathematically rigorous. How would you go about doing the algebra for this problem? Is there another approach that can be used?
 
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Use the fact that, for x and y positive, [itex]x^2> y^2[/itex] if and only if x> y.
[itex]a_j^2= \frac{j}{(j+5)^2}[/itex] and [itex]a_{j+1}^2= \frac{j+1}{(j+6)^2}[/itex].

Now use the fact that, for a, b, m, n all positive, a/b> m/n if and only if an> bm.
 
For j large, the terms are ≈ 1/√j , so the series of absolute values diverges.
 

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