A series connection (also called a series circuit) means placing electrical components one after another so they form a single loop (mesh). By the loop rule, the sum of the partial voltages equals the applied voltage.
Resistors in series
If $n$ ohmic resistors are connected in series to a source with voltage $U$, the partial voltages add to
$$
U = I R_1 + I R_2 + \dots + I R_n .
$$
Dividing by the constant current $I$ gives the total resistance:
$$
\frac{U}{I} = R_\mathrm{tot} = R_1 + R_2 + \dots + R_n
$$
and in general
$$
\boxed{R_\mathrm{tot} = \sum_{i=1}^n R_i}\,.
$$
Intuitively, putting resistors in series increases the effective conductor length, so the resistances simply add.
Capacitors in series
For capacitors in series, the effective plate separation increases, so the total capacitance decreases. Because the series current is the same everywhere, the total charge $Q$ on each capacitor is the same, and the voltages satisfy
$$
U = \frac{Q}{C_1} + \frac{Q}{C_2} + \dots + \frac{Q}{C_n} \,,
$$
which leads to
$$
\frac{1}{C_\mathrm{tot}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}
$$
or, in compact form,
$$
\boxed{\frac{1}{C_\mathrm{tot}} = \sum_{i=1}^n \frac{1}{C_i}}\;.
$$
The total capacitance is therefore smaller than the smallest individual capacitance. Note: after summing the reciprocals, take the reciprocal again to get $C_\mathrm{tot}$.
Inductors in series
Using the loop rule for inductors in series,
$$
U = L_1 \frac{\mathrm{d}I}{\mathrm{d}t} + L_2 \frac{\mathrm{d}I}{\mathrm{d}t} + \dots + L_n \frac{\mathrm{d}I}{\mathrm{d}t}\,.
$$
Since $\mathrm{d}I/\mathrm{d}t$ is the same everywhere in the loop, the total inductance is
$$
\boxed{L_\mathrm{tot} = \sum_{i=1}^n L_i}\;.
$$
Example
In three separate series circuits, find:
1) Total resistance of $100\,\Omega$ and $20\,\Omega$: $120\,\Omega$
2) Total capacitance of $50\,\mu\mathrm{F}$, $10\,\mu\mathrm{F}$, and $250\,\mathrm{nF}$: $243\,\mathrm{nF}$
3) Total inductance of $30\,\mathrm{mH}$, $120\,\mathrm{mH}$, and $240\,\mathrm{mH}$: $390\,\mathrm{mH}$