What series test would you use on these and why? I just started learning series , I am trying to put everything together...I have few some few random problems just to see what kind of strategy you would use here...


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*$\displaystyle\sum_{n=1}^\infty\frac{n^n}{(2^n)^2}$

*$\displaystyle\sum_{n=1}^\infty\frac2{(2n - 1)(2n + 1)}$;  telescoping series ?

*$\displaystyle\sum_{n=1}^\infty\frac1{n(1 + \ln^2 n)}$

*$\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n (n + 1)}}$;   integral test ? cause you could do u-sub?
 A: For the first i would use the root test, as this yields a pretty simple result. The ratio test will work too, just the calculation will be longer. In fact nearly every test should work for this one.
For the second I would use comparism test as it looks pretty much like 
$$\sum_{n=1}^\infty \frac{1}{n^2}$$
The root and the ratio test won't work here, as both would yield the result 1 which gives you nothing about convergence or divergence. Cauchy condensation could work but won't be so nice. Integral test will work too.
The third one is a bit more complicated. I would use comparism test or maybe cauchy condensation. (While the cauchy condensation some comparisn will be necessary).
For the fourth i would use comparism again as it looks pretty much like 
$$\sum_{n=1}^\infty \frac{1}{n}$$
Here ratio and root test won't work again, Cauchy condensation might work but will not be very trivial. 
A small remark for series number 2, with partial fracition decomposition you get a telescoping series with limit 1
$$\sum_{n=2}^\infty \frac{2}{(2n-1) (2n+1)} =\sum_{n=1}^\infty \frac{1}{2n-1} - \frac{1}{2n+1}=1$$
A: General and mixed hints:
$$\frac{2}{(2n-1)(2n+1)}=\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
$$\frac{n}{4}\xrightarrow[n\to\infty]{}\infty$$
$$\frac{2^n}{2^n(1+n^2\log^22)}\le\frac{1}{\log^22}\cdot\frac{1}{n^2}$$
A: #1 is just teeming with $n$th powers. Use the root test. #2 and #4 can be compared with $\sum 1/n^p$ for some $p$ (limit comparison). For #3 I would compare with $$\sum\frac{1}{n\ln n}$$ (which diverges by a standard application of the integral test).
