Inequality sign of sequences preserved in infinite convergence sums inequlities. Let $a_n$ and $b_n$ be two real positive number sequences, such that, for every $n\in \mathbb{N}$, $a_n&lt;b_n$. Also, let $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ be two convergent series. 
Why does the following inequality holds:

$$\sum_{n=1}^{\infty}a_n&lt;\sum_{n=1}^{\infty}b_n$$
Clearly $\sum_{n=1}^\infty a_n \leq \sum_{n=1}^\infty b_n$.
If you assume by contradiction that equality holds, then necessarily you must have $a_n = b_n$ for every $n$.

Btw, if $a_j\leq b_j$ for every $j$, to get the result it is enough to assume that there exists $m$ such that $a_m &lt; b_m$.

Namely, let $\delta := b_m - a_m &gt; 0$. If $n\geq m$ one has
$$
\sum_{j=1}^n a_j \leq \sum_{j=1}^n b_j - \delta,
$$
hence
$$
\sum_{j=1}^\infty a_j \leq \sum_{j=1}^\infty b_j - \delta.
$$
Clearly $\sum_{n=1}^\infty a_n \leq \sum_{n=1}^\infty b_n$.
If you assume by contradiction that equality holds, then necessarily you must have $a_n = b_n$ for every $n$.

Btw, if $a_j\leq b_j$ for every $j$, to get the result it is enough to assume that there exists $m$ such that $a_m &lt; b_m$.

Namely, let $\delta := b_m - a_m &gt; 0$. If $n\geq m$ one has
$$
\sum_{j=1}^n a_j \leq \sum_{j=1}^n b_j - \delta,
$$
hence
$$
\sum_{j=1}^\infty a_j \leq \sum_{j=1}^\infty b_j - \delta.
$$