real analysis

The limit exists for a), b), c).

It doesn't exist for d).

If justifications are required:
a) $\frac{n+1}{n^2+2}=\frac{1+1/n}{n+2/n}$ the denominator is monotonous in $n$ nad grows unboundedly, while the numerator is bounded. So the expression goes to $0$ as $n\to\infty$.
b) One has the same justification as a).
c) Note that
$|c_{n+1}-c_n| = |\frac{c_n+c_{n-1}}2-c_n| = \frac12 |c_{n}-c_{n-1}|$
so by induction
$|c_{n+1}-c_n| ≤ 2^{-n+1}|c_2-c_1|≤2^{-n+1}$
and then
$$c_n=\sum_{n=2}^N(c_{n+1}-c_n)+c_1$$
converges since the sum converges absolutely.
d) Note that $2n!≥3^n$ for $n$ large enough. Hence the individual summands of the series don't go to $0$ and the series cannot converge.