Answer check on two series I want to determine if these two are absolutely convergent, conditionally convergent or simply divergent.
1) $$\sum_{n=2}^\infty \left(\frac xn - \frac x{n-1}\right)$$
$$= \frac x2 - \frac x1 + \frac x3 - \frac x2 + \dots + \frac x{n+1} - \frac xn$$
$$=\frac x2 + \frac x{n+1}$$
$$=\frac{x(n+1)+2x}{2n+2}$$
$$=\frac{x(3+n)}{2n+2}$$
$$\lim \limits_{n\to\infty} \frac{x(3+n)}{2n+2} = \frac x2 $$
Hence it diverges $\forall x \ne 0$
$$\sum_{n=2}^\infty \left|\frac xn - \frac x{n-1}\right|$$
$$=\int_{2}^\infty \frac{|x|}{n^2 - n} dn$$
$$= \dots$$ Still working on it
2) $$\sum_{n=2}^\infty \left|\frac1{(\log n)^2} \right| \gt \left|\frac 1n \right|$$ Since $|\frac 1n |$ diverges as harmonic series, our series diverges.
 A: For (1): Don't split it up like that, the two series (independently) diverge (as you mentioned, they are harmonic). Instead, use the grouping of terms given to your advantage. Write a closed form expression for the partial sum (hint: telescoping). To determine absolute convergence, first combine the expression:
$$\left | \frac x n - \frac x {n-1} \right| = \frac{?}{n(n-1)}.$$
Now the question is does $\sum_{n=1}^\infty \frac{1}{n(n-1)}$ converge?
Finally, to address uniform convergence, what tests do you know? Specifically, I'm thinking of the Weierstrass M-Test, which could be used to show that the convergence is uniform on a compact set (i.e. a closed and bounded interval). The convergence is not uniform in general, however. (Why?)
For (2): Here the terms of the series are already positive, so conditional and absolute convergence are the same thing. Comparison test is the right idea, what you have written is false. You mean to write
$$\frac{1}{\log^2(n)}>\frac 1 n$$
for large enough $n$, and since $\sum_{n=1}^\infty \frac 1 n$ diverges, by the comparison test $\sum_{n=1}^\infty \frac 1 {\log^2(n)}$ diverges.
