Before answering the question let's make a note of few tests
1)Nth Term Divergence test:- For a series $\large\sum a_n$,
if the $\large\lim_{n->\infty}a_n \neq 0$ or DNE then the Series is a Divergent series.
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2)Limit comparsion test:- here it a comparision between two series, consider two series $\large\sum a_n$ and $\large\sum b_n$,
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$1.\large \lim_{n->\infty}\large\dfrac{a_n}{b_n} = C$, where C is constant $C>0$ then either both $\large\sum a_n$, $ \large\sum b_n$ converge or both diverge.
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$2.\large\lim_{n->\infty} \large\dfrac{a_n}{b_n} = 0$ and $\large\sum b_n$ converges then $\large\sum a_n$ also converges.
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$3.\large\lim_{n->\infty}\large\dfrac{a_n}{b_n} = inf$ and $\large\sum b_n$ diverges then $\large\sum a_n$ also diverges.
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3)P-series:- If a series of the form $\large\sum\dfrac{1}{n^p}$, these series converges whenever $p>1$ otherwise they diverges, one of the most famous series is harmonic series $\large\sum_{n=1}^{\infty} \large\dfrac{1}{n}$.
Now let's verify each option,
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$1.\large\sum sin(\large\dfrac{1}{n})$, By the divergence as n tends to infinity, $\large\dfrac{1}{n}$ will approach $\large0$ , $\large\sin (0) $ $ $is 0, by nth term divergence test we can’t conclude anything.
Lets try Limit comparison Test with $\large\sum b_n= \sum_{n=1}^{\infty} \dfrac{1}{n}$ and its a Divergent series.
$=\large\lim_{n->\infty}\large\dfrac{\sin(\dfrac{1}{n})}{\dfrac{1}{n}}$, ( $\dfrac{0}{0}$ form apply L’Hopital rule, then it will be)
$= \large\lim_{n->\infty}\large\dfrac{cos(\dfrac{1}{n}).(\dfrac{-1}{n^2})}{\dfrac{-1}{n^2}}$
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$=\large\lim_{n->\infty}\large\dfrac{cos(\dfrac{1}{n}).(\not{\dfrac{-1}{n^2})}}{\not{\dfrac{-1}{n^2}}}$
$=\large\lim_{n->\infty} \cos(\dfrac{1}{n})$
$=\large\cos(0) $
$=1$
Which is first condition of LCT , as the series $\sum b_n$ diverges same stays for $\sum a_n$, so given series also diverges.,
which means Option one is True.
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$2.\large\sum sin(\dfrac{1}{n^2})$, same as above, nth divergence test will be inconclusive and do limit comparision test with $\large\sum b_n=\large\sum \dfrac{1}{n^2}$, and value of
the $\large\lim_{n->\infty}\dfrac{sin(\dfrac{1}{n^2})}{\dfrac{1}{n^2}}$ $= 1$, so as $\large\sum b_n$ converges same as $\large\sum a_n$
so the second Option also true.
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$3.\large\sum \cos(\frac{1}{n})$, now as $n→\infty$, $\dfrac{1}{n}$ will be $0$ and $\cos(0)$ is $1$, so by nth term divergence test it will be divergent series.
so the Thrid option was also true.
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$4.\large\sum \cos(\dfrac{1}{n^2})$, now as $n \to \infty$, $\dfrac{1}{n^2}$ will be $0$ and $\cos(0)$ is $1$, so by nth term divergence test it will be divergent series, so the Answer is Option D.
