Divergence of Feynman diagram 
*

*Can we say whether the given Feynman diagram is divergent or not by just looking into the Feynman diagram?

*How to remove these divergences?
 A: Consider a generic Feynman diagram with,


*

*$L$ loops, $N_f$ number of internal fermion lines (or fermionic propagators) and $N_b$ number of internal boson lines (or bosonic propagators), 

*different kinds of vertices and the $i^{th}$ kind appears $N_{v_i}$ times, and

*number of derivatives in each vertex be $h_i$. 
Now suppose that we have written down the Feynman amplitude for this diagram. Let us see how it behaves when the internal momenta are in the UV region i.e. much larger than the masses of the particles in the loop. 


*

*Each internal fermion line contributes to a propagator which behaves like $\frac{1}{p}$, at large $p$, 

*Each internal boson line (vector or scalar) contributes to a propagator which behaves like $\frac{1}{p^2}$, at large $p$, and 

*Integration over each loop momentum contributes to $d$ powers of $p$ in the numerator due to the term $d^dp$, in $d$-dimensional spacetime. 
Therefore, the total power of $p$ in the diagram is therefore given by $$D=\text{power of $p$ in the denominator}-\text{power of $p$ in the denominator}\\=dL-N_f-2N_b+\sum\limits_{i}h_iN_{v_i}$$ $D$ is known as the superficial degree of divergence of the diagram. 
Naively, we expect a diagram to have a divergence proportional to $\Lambda^D$, where $\Lambda$ is the momentum cut-off for $D\neq 0$. Therefore, when $D<0$, the corresponding amplitude is expected to be convergent in the UV limit and for if $D>0$, the amplitude is expected to be divergent in the UV limit. With $D=0$, the diagram can have a logarithmic divergence i.e., $\log \Lambda$. For example, the vertex function in QED has $D=0$, and therefore will be logarithmically divergent. 
However, this naive expectation can often be wrong, for one of the following three reasons: (i) when a diagram contains divergent subdiagram, its actual divergence may be worse than that dictated by $D$, (ii) when symmetries (such as the Ward identity) cause certain terms to cancel, the divergence of a diagram may be reduced or even eliminated, and finally, (iii) a trivial diagram with no propagators and no loops has $D=0$ but has no divergence of any kind.
References
$1.$ An Introduction to Quantum Field Theory by Peskin and Schroeder.
$2.$ A First Course on Quantum Field Theory by Palash. B. Pal. 
A: Divergence: search for "superficial degree of divergence."
Removing or at least controlling the divergences: search for "renormalisation."
The latter topic is rich and deep and entirely non trivial. 
