# Real analysis: inequality limitsuperior/inferior

• K29
In summary, assuming a_n \geq 0 and that a_n+1\leq M\leq \sqrt[n]{|a_n|} \leq N\leq \overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_n|}
K29

## Homework Statement

Consider $\sum_{1}^{\infty} a_{n}, a_{n} \neq 0$

Show that

$\underline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}| \leq \underline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|}\leq \overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|)} \leq \overline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}|$

## The Attempt at a Solution

I have just been thinking about the limit superior part of the inequality:

$\overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|)} \leq \overline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}|$

Ok so I know that for all n
$|\frac{a_{n+1}}{a_{n}}| \leq M$
and that for all n
$\sqrt[n]{|a_{n}|} \leq N$

The best way I can see to prove that $N \leq M$ is to do a proof by contradiction
so if N>M then for a sufficiently large p such that:
$|\frac{a_{p+1}}{a_{p}}| \leq M \leq \sqrt[p]{|a_{p}|} \leq N$
I've spent a lot of time manipulating this equation, but I'm not getting to a contradiction.

Can anyone see a way to prove this or an alternative way to do the proof that works?

Another thing I tried was saying:
$\sqrt[n]{|a_{n}|} \leq N$
So
$a_{n} \leq N^{n}$
$\sqrt[n+1]{|a_{n+1}|} \leq N$
So
$a_{n+1} \leq N^{n+1}$

But I can't really do anything with that as far as I can see. Was just a thought.

Hi K29!

Let's get you started, will we.

First, we will change that annoying limsup by a limit. Note in general that

$\limsup{x_n}\leq \limsup{y_n}$

If for every subsequence $x_(n_k)$ that increases to the limsup, we have that

$$\lim{x_{n_k}}\leq \limsup{y_n}$$

In general, we can take the $x_n$ to converge (maybe to infinity) and we can take them to be increasing. So we need to show that

$$\lim{\sqrt[n]{|a_n|}}\leq \limsup{\frac{|a_{n+1}|}{|a_n|}}$$

Now, what if you take the logarithm of both sides?

Thanks for the help! Solved :)

## 1. What is the definition of limit superior/inferior in real analysis?

In real analysis, the limit superior (or upper limit) of a sequence is the largest number that the terms of the sequence eventually approach. Similarly, the limit inferior (or lower limit) of a sequence is the smallest number that the terms of the sequence eventually approach.

## 2. How is limit superior/inferior useful in real analysis?

Limit superior/inferior allows us to determine the behavior of a sequence as it approaches infinity. It also helps us prove convergence and divergence of sequences, as well as identify accumulation points.

## 3. Can limit superior/inferior be equal to the actual limit of a sequence?

Yes, the limit superior/inferior can be equal to the actual limit of a sequence. This occurs when the sequence is eventually constant, meaning that all terms after a certain point are equal to the limit.

## 4. How is limit superior/inferior calculated?

To calculate limit superior, we take the supremum (or the smallest upper bound) of the set of all subsequential limits of a sequence. To calculate limit inferior, we take the infimum (or the largest lower bound) of the set of all subsequential limits of a sequence.

## 5. Can limit superior/inferior be used for infinite series?

Yes, limit superior/inferior can be used for infinite series. In this case, the limit superior/inferior represents the largest and smallest possible values that the partial sums of the series can approach.

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