Finding counterexamples for infinite series.

Give either a justification or a counterexample for each of the following statements about real series.

$\quad(a)$ If $ka_k\to0$ as $k\to\infty$, then $\sum a_k$ converges.

$\quad(b)$ If $\lim\limits_{k\to\infty}\frac{a_{k+1}}{a_k}$ exists and equals $L>1$, then $\sum a_k$ diverges.

$\quad(c)$ If $\sum a_k$ converges and $\lim\limits_{k\to\infty}\frac{a_k}{b_k}=1$, then $\sum b_k$ converges.

I think all propositions are false. But struggling to find counterexamples. Is there some sort of systematic way of constructing counterexamples for these sorts of series? Can someone explain their thinking process in giving counter examples for the above?

Solution 1:

a. As shown in the comment, $a_k = 1/(k \log k)$ fails.

b. True, since $\newcommand{\abs}[1]{\left\vert #1 \right\vert} \newcommand\rme{\mathrm e} \newcommand\imu{\mathrm i} \newcommand\diff{\,\mathrm d} \DeclareMathOperator\sgn{sgn} \renewcommand \epsilon \varepsilon \newcommand\trans{^{\mathsf T}} \newcommand\F {\mathbb F} \newcommand\Z{\mathbb Z} \newcommand\R{\Bbb R} \newcommand \N {\Bbb N} \renewcommand\geq\geqslant \renewcommand\leq\leqslant \newcommand\bm\boldsymbol \abs {a_{k+1}}/\abs{a_k} > (1+c)$ for some $c >0$ for all large $k$, and eventually $\abs {a_k} >M(1+c)^{k} \to +\infty$, hence the series cannot converges.

In detail, by the limit equation and the definition, there is some $N\in \N^*$ s.t. whenever $k \geq N$, $$ \frac {\abs {a_{k+1}} }{\abs {a_k}} > \frac {1+L}2 > 1, $$ hence $$ \abs {a_{k+1}} = \abs {a_N} \frac {(1+L)^{k-N}}{2^{k-N}} [k \geq N], $$ then taking the limit $k \to +\infty$ we get $\abs {a_k} \to +\infty$. The convergence of $\sum x_k$ requires $\lim_{k \to +\infty} x_k =0$, since if $S = \sum {x_k}$, then $x_k = \sum_1^k x_j - \sum_1^{k-1} x_j$ and $\lim x_k = S - S = 0$. The deduction above shows $a_k$ does not meet this requirement, hence it diverges.

c. False. $$ a_n = \frac {(-1)^n}{\sqrt n}, b_n = a_n + \frac 1n. $$