How do i determine if a sequence converges or diverges

[ineedhelpnow]

like these problems for example. if it converges then I am supposed to find the limit.

$a_n=\frac{\sin\left({2n}\right)}{1+\sqrt{n}}$

$\left|\sin\left({2n}\right)\right| \le 1$
$a_n=\frac{(-3)^n}{n!}$

$\left|a_n\right| > 0$

 

[evinda]

like these problems for example. if it converges then I am supposed to find the limit.

$a_n=\frac{\sin\left({2n}\right)}{1+\sqrt{n}}$

$\left|\sin\left({2n}\right)\right| \le 1$

We know that : $|\sin(2n)| \leq 1 \Rightarrow -1 \leq \sin(2n) \leq 1 \Rightarrow -\frac{1}{1+ \sqrt{n}} \leq \frac{\sin(2n)}{1+ \sqrt{n}} \leq \frac{1}{1+ \sqrt{n}}$

Now,we take the limit $n \to +\infty$ and we have:

$$\lim_{n \to +\infty} -\frac{1}{1+ \sqrt{n}} \leq \lim_{n \to +\infty} \frac{\sin(2n)}{1+ \sqrt{n}} \leq \lim_{n \to +\infty} \frac{1}{1+ \sqrt{n}} \Rightarrow 0 \leq \lim_{n \to +\infty} \frac{\sin(2n)}{1+ \sqrt{n}} \leq 0$$

So,according to the Squeeze theorem:
$$\lim_{n \to +\infty} \frac{\sin(2n)}{1+ \sqrt{n}}=0$$

Therefore,the sequence converges to $0$.

 

[ineedhelpnow]

so if it were $\infty$ it would diverge?

 

[evinda]

so if it were $\infty$ it would diverge?

Yes..

 

[ineedhelpnow]

if you take the limit of the second one you get 0. so converges?

 

[evinda]

if you take the limit of the second one you get 0. so converges?

$$\frac{3^n}{n!}=\frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{3}{5} \cdots \frac{3}{n} \leq \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdots \frac{3}{4}=3 \cdot (\frac{3}{4})^{n-3} $$

$$ \text{As } n \to +\infty , (\frac{3}{4})^{n-3} \to 0 \text{ and } \frac{3^n}{n!}>0 \ \text{ so,according to the Squeeze theorem: } \frac{3^n}{n!} \to 0$$

 

[ineedhelpnow]

why is it to 3/4

 

[ineedhelpnow]

never mind. i understand

 

[evinda]

why is it to 3/4

We take this number,because we know that: $\displaystyle{ a^n \to 0 }$,as $\displaystyle{n \to +\infty }$,when $|a|<1$ and $\frac{3}{4}$ satisfies this condition..

 

[Prove It]

so if it were $\infty$ it would diverge?

Only if you can show that your expression is GREATER than the quantity that $\displaystyle \begin{align*} \to \infty \end{align*}$...

 

[ineedhelpnow]

what do you mean?

 

[Deveno]

Basically, the idea is to compare two sequences "term-by-term".

One caveat, one may have to go out quite a few terms to get "dominance" of one sequence over another. But discarding any finite amount of "initial terms" does not affect the behavior of the sequence "at infinity".

Now if for some $N$, we have for EVERY $n > N$, $b_n > a_n \geq 0$, and the sequence $\{b_n\}$ converges, then the $a_n$ are "trapped" between the limit of the $b_n$ and 0 (as $n$ "approaches infinity", that is to say, is sufficiently large).

This, by itself, doesn't mean $\{a_n\}$ converges, it might just "oscillate wildly" between the limit for the $b_n$ and 0.

If, however, the $b_n$ converge to 0, the $a_n$ have no choice in the matter.

Now if the $a_n$ fail to converge, the behavior of the $b_n$ will be "even worse" (it can't get better, because the $a_n$ are "keeping the $b_n$ away" from any potential limit).

Calculus teachers (sly devils that they are), don't want to make this easy for you, so they typically want you to look at:

$a_n = \dfrac{c_n}{d_n}$

where both "top and bottom" behave the same "at infinity" (both go to 0, or both become infinitely large), and it becomes a matter of "practice" recognizing which "side" wins, and often, you have to resort to some OTHER sequence for comparison.

I wish I could tell you "which other sequences" to use, but this is almost ALWAYS done on a case-by-case basis.

Of course, teachers "rig the game", they assign problems THEY already know the answers to. In actual practice, it can be VERY DIFFICULT to decide if a sequence converges. It's an art-form, really, there are a few "tricks" to use, but sometimes everything you try doesn't help.

 

[ineedhelpnow]

so it can be $\infty$ at times and still be convergent?

 

[Deveno]

so it can be $\infty$ at times and still be convergent?

You'll have to be more specific, I'm afraid, as I have no idea what you mean by this.

Try to avoid speaking of infinity as if it was a number-it's not. It is sometimes used as a handy abbreviation for more wordy ways of saying things.

For example, when we say a sequence "goes to infinity", we don't mean that:

There exists $k$ such that: $a_k = \infty$.

We mean instead: for every natural number $N$, there IS a $k$ such that for all $n > k$, we have $a_n > N$.

Note the last statement doesn't even mention infinity, it just means that there is no $N$ that "bounds" the sequence $\{a_n\}$.

For example, if:

$a_n = \dfrac{1}{n - 3}$

then $a_3$ is undefined, it's not "infinity".

 

[ineedhelpnow]

if i am given a sequence and i take the limit and it goes to infinity, does that mean it diverges. Likewise, if I take the limit of a sequence and i get a real number (finite number) such as 0 or 1, that means it converges. right?

 

[MarkFL]

If the $n$th term increases/decreases without bound, then you know it diverges, but in order for the series to converge the $n$ term must approach zero as $n\to\infty$, and while this is a necessary condition, it is not a sufficient condition...it must approach zero quickly enough for the series to converge. Consider the harmonic series:

\(\displaystyle S=\sum_{k=1}^{\infty}\frac{1}{k}\)

The $k$th term does indeed approach zero as $k$ increases without bound, but the series itself diverges.

 

[Prove It]

If the $n$th term increases/decreases without bound, then you know it diverges, but in order for the series to converge the $n$ term must approach zero as $n\to\infty$, and while this is a necessary condition, it is not a sufficient condition...it must approach zero quickly enough for the series to converge. Consider the harmonic series:

\(\displaystyle S=\sum_{k=1}^{\infty}\frac{1}{k}\)

The $k$th term does indeed approach zero as $k$ increases without bound, but the series itself diverges.

Except this thread is about convergence of SEQUENCES, not series :P

 

[ineedhelpnow]

yeah I am talking about if I am given like $a_n$ or something and then i asked if it converges or diverges.

 

[MarkFL]

yeah I am talking about if I am given like $a_n$ or something and then i asked if it converges or diverges.

Then just take the limit...if it is a real number, it converges. :D

 

[ineedhelpnow]

excellent answer Mark. very clear answer to my question :)
appreciate all the other answers too :)