What is $sup{ (-1)^n + 1/n} $? My doubt is that $\sup (A+B) = \sup A + \sup B$.
So, if we consider  $A =(-1)^n$ and $B=\frac{1}{n},$
then $\sup A = 1 $  and   $\sup B = 1$.
So will $\sup (A+B) = \sup{ (-1)^n + 1/n}  = \sup (-1)^n   +  \sup \frac{1}{n} = 1+1 =2 ?$
Is this correct or have I misunderstood somewhere please explain?
Also could you please provide the correct method to solve the above problem?
 A: The problem here is that in the statement
$$
\sup (A + B) = \sup A + \sup B
$$
the definition of "$+$" is given by 
$$
A+B=\{a+b\mid a\in A, b\in B\},
$$
and $A$ and $B$ are sets (of real numbers)
. 
You, in trying to apply it, appear to be adding two sequences termwise, i.e., you're defining a new sequence 
$$
(A+B)_n = A_n + B_n
$$
With this definition of addition (and the implied notion that "sup" of a sequence is the sup of the set of items in the sequence) the "theorem" is no longer a theorem, as your counterexample shows. 
A: You cannot use this property. This is because were are not talking about elements in sets, but about sequences which depend on the same $n$, therefore, these sequences are not 'disjoined'. 
An easy way to look at this problem is that you want to "maximize" both $(-1)^n$ and $1/n$ simultaneously. Notice that both are bounded by $1$, so we want to show that we can find $n \in \mathbb{N}$ such that both components are equal to one. Notice that for $(-1)^n=1$, we must have that $n$ is even, i.e. $n=2k$. Substituting the later into the second component, we have $\frac{1}{n} = \frac{1}{2k}$, where $\sup\big({\frac{1}{2k}}\big) = \frac{1}{2}$, where $k=1$, and thus $n = 2k = 2$. Hence, 
\begin{equation}
\sup\bigg(\frac{1}{n} + (-1)^n\bigg) = \frac{1}{2} + 1 = 1\frac{1}{2}
\end{equation}
A: $\newcommand{\s}{: n\in\mathbb{Z}^{+}}$ If $A = \{ (-1)^n \s \}$ and $B = \{ 1/n \s \}$, it is not true that $\{ (-1)^n + 1/n \s \}$ equals $A+B$. This is because $A+B$ refers to the set of all sums of pairs of elements of $A$ and $B$ where the "$n$" can be different for each. That is, $A+B = \{ (-1)^{n_1} + 1/n_2 : n_1, n_2\in \mathbb{Z}^{+}\}$, whereas in $\{ (-1)^n + 1/n \s \}$, the "$n$" is the same in both terms.
For example, $(-1)^3 + 1/2 = -1/2$ is in $A+B$ but you can show that $-1/2$ is not in $\{ (-1)^n + 1/n \s \}$.
Thus, your reasoning was unfortunately incorrect.
(Written using functions, you can say that $\sup\limits_n (f(n)+g(n))$ in general does not equal $\sup\limits_n f(n)+\sup\limits_n g(n)$.)
