A: if $\sum a_n$ converges, then so does $\sum a_n^4$.,
before concluding whether this is true or not let’s do some analysis on these kind of statements,
now whenever a series $\sum a_n$ is convergent it means $\exists N $ such that for $\forall n\geq N$, $|a_n|<1$, which means after some point the values in series $\sum a_n$ comes closer and closer to 0, so now lets consider two statements,
observation 1:- $\sum a_n$ is convergent, as i mentioned above, $\exists N$ such that $\forall n \geq N $ , $|a_n| < 1$
statement 1$:$if $\sum a_n$ converges then $\sum a_n^2 $ also converges.
statement 2$:$ if $\sum a_n$ converges then $\sum a_n^3$ also converges.
which of them you think its correct?, the answer is first one is incorrect and second one is correct.
consider this series $\sum a_n=\sum (-1)^{n+1} \frac{1}{n^{1/2}}$ so $\sum a_n^2 = \sum \frac{1}{n}$,
series $a_n$ is a convegent and $a_n^2$ is a divergent, this is happening because you cant conclude thats whether $\forall n \geq N $ $a_n^2 \leq a_n$, because $\sum a_n$ consists of negative terms and those terms square will be greater than the $a_n$, so that’s the convergence so $\sum a_n^2$ is not dependent on $\sum a_n$,
consider this series $\sum a_n = \sum (-1)^{n+1} \frac{1}{n}$ , then $\sum a_n^2 = \sum \frac{1}{n^2}$ and both these series are convergent.
but its true that $a_n^2 \leq |a|$, now if $\sum |a_n| $ is convergent series then we can say that $a_n^2 \leq |a|$ $\forall n \geq N $, by this i can say that,
$\sum_{m=N}^{inf} a_n^2 $ $\leq $ $\sum_{m=N}^{inf} |a_n| $
so now by direct comparision test we can conclude that $\sum a_n^2$ is convergent series.
so option B is true , by the same process as above you can verify why statement 2 is correct and option A is false. (Hint :- think about $a_n^3 \leq a_n $ and $a_n^4 \leq a_n$)
coming to option C:- we can go with counter examples,
$\sum a_n$ = $\sum \frac{1}{n^{2/3}}$ which is a divergent, because its p-series of value p<1
$\sum an_n^3$ = $\sum \frac{1}{n^2}$, which is a convergent series
so option C is false.
Option D:-
$\sum a_n$ = $\sum (-1)^{n+1} \frac{1}{n}$, this series is condinally convergent that is $\sum a_n$ converges and $\sum |a_n|$ diverges.
$\sum a_n^2$ = $\sum \frac{1}{n^2}$ which is convergent series so option D also false
the only correct option is OPtion B .
