If \(a_n\) = some expression in n, just put the values of n in the expression to find the terms. 

In this case \(a_n = \dfrac{2^{n}}{1+3^{n+1}}\)

\(\therefore\;\;a_1 = \dfrac{2^1}{1+3^2} = \dfrac{2}{10} = \dfrac{1}{5}\)

\(a_2 = \dfrac{2^2}{1+3^3} = \dfrac{4}{28} = \dfrac{1}{7}\)

\(a_3 = \dfrac{2^3}{1+3^4} = \dfrac{8}{82} = \dfrac{4}{41}\)

\(a_4 = \dfrac{2^4}{1+3^ 5} = \dfrac{16}{244} = \dfrac{4}{61}\)

\(a_5 = \dfrac{2^5}{1+3^6} = \dfrac{32}{1+729} = \dfrac{32}{730} = \dfrac{16}{365}\)

Observe the pattern of numerator and denominator of the term and try to relate them to the term number.

In this case 

\(a_1 = \dfrac{1}{2}, a_2=\dfrac{-4}{3} , a_3= \dfrac{9}{4}, a_4= \dfrac{-16}{5}, a_5= \dfrac{25}{6}\)

Observe that the denominator starts with 2 in \(a_1\), and increases by 1, whereas numerator is a perfect square of consecutive numbers with alternate positive and negative signs.  

\(\therefore\;\; a_n = \dfrac{(-1)^{n+1} \; n^2}{n+1}\)

A sequence is said to be increasing if \(a_n < a_{n+1}\) for all \(n \geq 1 \; (n \in N)\).

Check the options 

(a)     \(a_n = \dfrac{1}{5n+4}\)

        \(a_n < a_{n+1}\)

\(\Rightarrow \; \dfrac{1}{5n+4}\; < \dfrac{1}{5(n+1) +4}\)

or \(\dfrac{1}{5n+4}<\dfrac{1}{5n+9} \) or \(5n+ 4 > 5n +9 \)

or \(4>9\) which is not true.

\(\therefore \; a_n\) is a decreasing sequence .

(b)   \(a_n = (-3)^{n+1}\)

         \(a_n < a_{n+1}\)

\(\Rightarrow \;(-3) ^{n+1} < (-3) ^{n+2}\)

\(\Rightarrow \; (-1)^{n+1} × 3^{n+1}\;<\; (-1)^{n+2} × 3^{n+2}\)

or  \((-1)^{n+1} < (-1) ^{n+2} × 3\)

which may or may not be true according to n.

\(\therefore\; a_n \) is non monotonic .  

(c)  \(a_n = n + \dfrac{1}{n}\)

       \(a_n <a_{n+1}\)

\(\Rightarrow \; n+ \dfrac{1}{n } < n+1 + \;\dfrac{1}{n+1}\)

or    \(\dfrac{1}{n} < \;1 + \dfrac{1}{n+1} \; or \dfrac{n+1+1}{n+1} - \dfrac{1}{n} > 0\)

\(or \;\; \dfrac{n^2 + 2n -n -1}{n (n+1)} > 0 \;\; or \;\; \dfrac{n^2 + n -1}{n (n+1)} > 0\)

which is true.

\(\therefore\;\; a_n \) is an increasing sequence.

(d)    \(a_n = \dfrac{1}{2n+5}\)

         \(a_n<a_{n+1}\)

\(\Rightarrow \; \dfrac{1}{2n+5}<\dfrac{1}{2 (n+1) +5} \; or \; \dfrac{1}{2n+5} < \dfrac{1}{2n+7}\)

\(or\;\; 2n+5 > 2n+7\;\; or\;\; 5>7 \)  which is not true. 

\(\therefore \; \; a_n \) is a decreasing sequence.  

For a convergent sequence 

\(\lim\limits _{n\to \infty} \; a_n\; = L \)    where   L is a finite number, called the limit of the sequence .  

In this case 

\(L = \lim\limits _{n \to \infty } \sqrt{\dfrac{5n + 3}{2n + 5}}\)

\(= \lim\limits _{n\to \infty} \sqrt {\dfrac{5 + \dfrac{3}{n}}{2 + \dfrac{5}{n}}}\)  ( Divide Numerator & Denominator by n in the square root )

\(= \sqrt {\dfrac{5+0}{2+0}} \;\;\;=\;\;\;\sqrt{\dfrac{5}{2}}\)

\(\therefore \;\;L = \sqrt {\dfrac{5}{2}}\)

For a convergent sequence 

\(\lim\limits _{n \to \infty} \; a_n\; = L \)    where   L is a finite number, called the limit of the sequence.   

In this case 

\(a_n = ln (3n^2 +1) - ln (n^2 +4)\)

\( = ln \left(\dfrac{3n^2 + 1}{n^2 +4}\right)\)        ( use ln a – ln b = \(ln\dfrac{a}{b}\) )

\(\therefore\) If \(a_n = f (n)\) then \(f(n) =ln \left(\dfrac{3n^2 + 1}{n^2 + 4}\right)\)

\(\therefore \; L = \lim\limits _{n\to \infty }\; ln \left(\dfrac{3n^2 + 1}{n^2 + 4}\right)\)

\(= \lim\limits _{n\to\infty } \; ln \left(\dfrac{3+\dfrac{1}{n^2}}{1+\dfrac{4}{n^2}}\right)\; \)

\(= ln \;\dfrac{3+0}{1+0}\\ = ln\; 3\)