Sequences and Series Comparison

Question

During a lecture, my lecturer said that the following is possible

$a_n < b_n$

but

$\sum_{n=0}^\infty a_n \leq \sum_{n=0}^\infty {b_n} $

However, I am not able to come up or find any examples.

Help?

What's wrong with the following line of thought?

$a_n < b_n$

$\sum_{n=0}^N a_n < \sum_{n=0}^N {b_n} $

$\lim_{n \to \infty} \sum_{N=0}^N a_n < \lim_{n \to \infty}\sum_{N=0}^N {b_n} $

$\sum_{n=0}^\infty a_n < \sum_{n=0}^\infty {b_n} $

Answer 1

That is impossible unless the series are divergent to plus or minus $\infty$.

New: added some justification.
Denote $A_n := \sum_{k=0}^n a_k$ and $B_n := \sum_{k=0}^n b_k$.
Note that $$B_n - A_n = \sum_{k=0}^n (b_k-a_k) > \sum_{k=0}^{n-1} (b_k-a_k) > \cdots > b_0-a_0 > 0.$$ Thus the gap between the partial sums increases and is always greater than $b_0-a_0>0$.
Now if $A_n\to A$ and $B_n\to B$, where $A,B$ are real numbers, then for sure $B-A\geq b_0-a_0>0$.
However, $A_n\to\infty$ and $B_n\to \infty$ is also possible if, e.g., $a_n\equiv 1$ and $b_n\equiv 2$.

Answer 2

For two sequences $\{a_n\}$ and $\{b_n\}$, if $a_n < b_n \forall n \in N$, then necessarily $\sum_{n=1}^{\infty}a_n \lt \sum_{n=1}^{\infty}b_n$, assuming both series converge.

As a motivation for the proof, consider this: there are two glasses $A$ and $B$. You poured $2$ units of water into $A$ and $3$ units into $B$. Initially, $B$ has $1$ unit more water. Then you keep adding water to both glasses, but always add a bit more to $B$ than to $A$. You can then be assured that there will always be at least $1$ unit difference in water levels. If the glasses never overflow (both series converge), their final levels also differ by at least $1$ unit.

The proof, as it turns out, is quite trivial. Consider $$c_n = b_n - a_n$$ Clearly, $c_n \gt 0 ~\forall n$. Then, $$\sum_{n=1}^{N}c_n > c_1$$ meaning the partial sum sequence has a limit that is at least $c_1$. Assuming that $\sum_{n=1}^{\infty}a_n = a_0$ and $\sum_{n=1}^{\infty}b_n = b_0$, $b_0 - a_0 > c_1$.

That said, I would like to mention three things.

1. Since $x > y \implies x \ge y$, the statement your lecturer made is not technically wrong. He/she could have made the stronger statement containing the strict inequality, but this is not wrong.
2. Taking limit on both sides may not preserve inequality. While it is true that $\frac{1}{n+1} \lt \frac{1}{n}~ \forall n\in N$, the limits of both sequences are $0$.
3. A limit might not even exist in the first place. For example, $-n < \frac{1}{2^n} ~ \forall n$ but the left limit doesn't even exist.

Answer 3

Confusion:

Let $s_m \lt r_m$, $m\in \mathbb{N}, $

convergent sequences.

Then

$\lim_{n \rightarrow \infty} s_n \le \lim_{n \rightarrow \infty}r_n.$

The limits could(!) be equal.

Example:

$s_n = 1-1/n;$ $r_n = 1+1/n.$

Note: a sequence $s_n$ can be a partial sum.