if$ x_n If i have two sequences $x_n$ and $y_n$, where$\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ both converge, and for all $n
\in \mathbb{N}$ $x_n < y_n$, can I conclude that $$\sum_{n=1}^\infty x_n < \sum_{n=1}^\infty y_n?$$
or is that $$\sum_{n=1}^\infty x_n \leq \sum_{n=1}^\infty y_n?$$
 A: Hint: If $y_n > x_n$ for each $n$, then $y_n=x_n + \varepsilon_n$, where $\varepsilon_n = y_n - x_n> 0$ for each $n$. Then $$\sum y_n=\sum (x_n + \varepsilon_n) = \sum x_n + \sum \varepsilon_n$$ 
A: The inequality can be made strict. To see this, let $A_N=\sum_{n=2}^{N} x_n$ and $B_N=\sum_{n=2}^N y_n$ for every integer $N\geq 2$. Since strict inequalities are preserved by finite sums, $A_N<B_N$ for all integers $N\geq 2$, and, since strict inequalities between the members of two convergent sequences are preserved as a weak inequality (in general) in the limit,
$$\sum_{n=2}^{\infty} x_n=\lim_{N\to\infty} A_N\leq\lim_{N\to\infty} B_N=\sum_{n=2}^{\infty} y_n.$$
Now comes the part that makes the weak inequality actually strict. Given that $\sum_{n=2}^{\infty} x_n\leq\sum_{n=2}^{\infty} y_n$ and $x_1<y_1$, the conclusion is that
$$\sum_{n=1}^{\infty} x_n=x_1+\sum_{n=2}^{\infty} x_n<y_1+\sum_{n=2}^{\infty} y_n=\sum_{n=1}^{\infty} y_n.$$
A: No, from what data you provided, there is no way to conclude anything about the two sums, and here is why: 


*

*Yes bot $\sum x_n$ and $\sum y_n$ converge but to what values? This is the first problem that occures and most trivial.

*Looking at the previous answer, it's a solid idea to look for a difference $x_n - y_n$ for each $n$ but then again the sum of those differences might not converge, which makes it useless.
