Inequalities at the level of infinite series This question is tangentially related to a previous question I asked regarding proving that real numbers have at most two decimal expansions, but it involves more real analysis, so I thought it would be best to post a new question.
For finite sums, there is no issue with term-by-term inequalities. But what about infinite sums? For example, suppose I have a sum
$$
\sum_{i=1}^{\infty} a_i x_i
$$
where $a_i \leq 9$ for all $i$. Can I then assert
$$
\sum_{i=1}^{\infty} a_i x_i \leq \sum\limits_{i=1}^{\infty} 9 x_i.
$$
What if add a nonnegative quantity $C$ on the right-hand side. Can I assert that
$$
\sum_{i=1}^{\infty} a_i x_i \leq \sum_{i=1}^{\infty} a_i x_i + C?
$$
What if that $C$ is another infinite sum?
My intuition is that this is allowed provided that the sums in question exist and are absolutely convergent. Is this correct?
 A: I believe that your problem is more about really understanding the proof than about merely proving the results (or reciting the proof in the textbook). I had been confused by the idea of moving to infinity as well when I first studied analysis, so I’ll try to be more explanatory than formal here.
Note that infinite series is defined as the limit of partial sums. As you already see, “for finite sums, there is no issue with term-by-term inequalities.” Therefore, if given $a_i\leq b_i$ for all $i\in\mathbb N$, you can first establish inequality between corresponding (finite) partial sums:
$$\sum_{i=1}^n a_n\leq \sum_{i=1}^n b_n,\ \forall n\in\mathbb N$$
Let $A_n=\sum_{i=1}^n a_n$ and $B_n=\sum_{i=1}^n b_n$. Then we know
$$A_n\leq B_n$$
And the inequality
$$\sum_{i=1}^\infty a_n\leq \sum_{i=1}^\infty b_n$$
can be rewritten as
$$\lim_{n\rightarrow\infty}A_n\leq\lim_{n\rightarrow\infty}B_n$$
Now let’s forget that $A_n$ and $B_n$ are “aliases” of sums and rather just treat them as numbers. The remaining work is to show that $A_n<B_n$ guarantees $\lim\limits_{n\rightarrow\infty}A_n\leq\lim\limits_{n\rightarrow\infty}B_n$.
At this point of time we need to resort to the definition of limit. Suppose $\lim\limits_{n\rightarrow\infty}A_n=A$ and $\lim\limits_{n\rightarrow\infty}B_n=B$. Further suppose for contraction that $A>B$. Then for any $\varepsilon>0$ there exists $N\in\mathbb N$ sufficiently large such that $n>N$ guarantees
$$A-\varepsilon<A_n<A+\varepsilon$$
and
$$B-\varepsilon<B_n<B+\varepsilon$$
In particular, if we take $\varepsilon=\frac12(A-B)$, then
$$B_n<B+\frac12(A-B)=\frac12(A+B)=A-\frac12(A-B)<A_n$$
which contradicts our assumption that $A_n<B_n$.
Informally speaking, it may be counterintuitive that $A$ could be equal to $B$, but $A$ cannot be so outrageous as to be greater than $B$, or otherwise some terms $A_n$ will be taken by $A$ to the right side of $B_n$.
Thus we have established that
$$A=\lim_{n\rightarrow\infty}A_n\leq B=\lim_{n\rightarrow\infty}B_n$$
i.e.
$$\sum_{i=1}^\infty a_n\leq \sum_{i=1}^\infty b_n$$
This immediately answers your first question when you plug in the values.
Your second question is almost trivial—there’s absolutely no problem if you can keep in mind that a infinite sum as a whole is nothing more than a number, just like $\pi$ is nothing more than a number, for which you can make such conclusions like $\pi<\pi+C$ provided $C>0$, even though you don’t know the precise value of $\pi$.
I guess, though, that your next step is to somehow mix the terms of the two infinite sums on the right hand side—if $C$ is another infinite sum—together. In that case you need to be more careful and may need something like absolute convergence.
