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.

• You can not distribute the $\sum_n$ in the first series. You can sum it using telescoping technique. – Mhenni Benghorbal Jun 26 '14 at 1:32
• Overall, your notation is incorrect in both examples. I have some idea of what you mean, but it is not what you have written. – process91 Jun 26 '14 at 1:36
• @MhenniBenghorbal But the series of a sum is the sum of the series? – Tony Jun 26 '14 at 1:45
• The term is "diverges" not "diverges absolutely." – André Nicolas Jun 26 '14 at 1:45
• @user151558 Only if the series separately converge, these do not. – process91 Jun 26 '14 at 1:46

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?
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.
• Oh the past exam question says afterwards: "For the series in part (a)(here 1)) above, is the convergence uniform in $x$". That is why I wrote that in there – Tony Jun 26 '14 at 1:50
• So the closed form here is $|\frac{-x}{n^2 - n}|$ (for absolute convergence I mean) – Tony Jun 26 '14 at 1:54
• @user151558 The "closed form" was for the conditional convergence. Yes, we find: $$\sum_{n=1}^\infty \left|\frac x n - \frac x {n-1} \right| = \sum_{n=1}^\infty \frac{|x|}{n(n-1)},$$ Now does this converge or diverge? (One test that would work would be the comparison test. The integral test also would work.) – process91 Jun 26 '14 at 1:57