Learning to use the Cauchy criterion for infinite series

[Hall]

Homework Statement

Determine if the series $\sum_{n=1}^{\infty} \frac{2^n}{n !}$ converges.

Relevant Equations

Partial sums will be denoted by $s_n$.

$s_1=2$
$s_2=4$
$s_3=5.333$
$s_4=5.9999$
$(s_n)$ is increasing, but unable to guess a bound. Let's see if Cauchy criterion can do something.

For n>2,
$$
s_{n+k} - s_n = \frac{2^{n+1} }{(n+1)!} + \frac{ 2^{n+2} }{(n+2)!} + \cdots \frac{2^{n+k} }{(n+k)!}
$$
$$
s_{n+k} - s_n < \frac{2^{n+1} }{ (n+1)!} + \frac{ 2^{n+2} }{n (n+1)!} + \frac{ 2^{n+3} }{n^2(n+1)!} + \cdots \frac{2^{n+k} }{n^{k-1}(n+1)!}
$$
The RHS is a Geometric series with $a = \frac{2^{n+1} }{ (n+1)!}$ and $r = \frac{2}{n}$. So, by sum formula of Geometric series (as r is less than 1, coz we assumed n >2, therefore series converges)
$$

s_{n+k} - s_n < \frac{2^n}{(n+1)!} \frac{n^k - 2^k}{n^{k-1} } ~\frac{1}{n-2} \lt \frac{2^n}{(n+1)!} \frac{(n-2)~k~n^{k-1}}{n^{k-1}}= \frac{2^n}{(n+1)!} k
$$

Am I correct so far? My doubt is though $\frac{2^n}{(n+1)!}$ can be made as small as we please by taking large n, but won't taking a far off term make $k$ very big, thus, nullifying the nulling effect of $\frac{2^n}{(n+1)!}$?

 

[Office_Shredder]

Did someone tell you to use the cauchy criterion here? The ratio test is just better.

Also you did something bad. You can bound a finite geometric series by an infinite one, which has sum $1/(1-r)$. That should let you get the result you want, and it won't depend on k at all.

 

[Hall]

Did someone tell you to use the cauchy criterion here? The ratio test is just better.

I have not learned ratio test yet. As I'm learning totally by reading books, I have found the way to learn each criterion of convergence only when the previous ones fail. Like in this case the monotone sequence method has failed, so I tried the next one, i.e., Cauchy's Criterion.

Also you did something bad. You can bound a finite geometric series by an infinite one, which has sum 1/(1−r). That should let you get the result you want, and it won't depend on k at all.

That's a very nice thought. So,
$$
s_{n+k} - s_n \lt \frac{2^{n+1}}{(n+1)!} ~\frac{ n}{n-2}
$$
Do I need to show that limit of RHS is 0?

 

[Office_Shredder]

I feel like you already did, since you handled the left half. The right half just goes to 1.

 

[mathwonk]

you are using criteria designed for sequences. series have specially suited criteria, which are easier to apply, usually called "comparison" tests. look for those.

 

[Hall]

@Office_Shredder As the following doubt is very much related to this one, and starting a new thread for that little doubt doesn't seem very reasonable, so, I'm asking it here.

I read a new method for checking the convergence, called Cauchy Condensation test, which says
"the series $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=1}^{\infty} 2^n a_{2^n}$ converges". I would like to know can we somehow say where to apply this Condensation test? The only way to know if this test would work or not is to apply it and after doing all laborious work to come back and say "Ah! it doesn't work". My question has gravity, because you see Ratio test is quite simple to apply, Cauchy sequence test is also, somewhat, easy to check; in these two methods we get an idea within 2 or 3 steps if we're in right direction, but that condensation test is used, I guess, only for the series which somehow involve powered harmonic series or some multiplication of it.

 

[Office_Shredder]

You haven't stated the condensation test correctly. The sequence needs to be non negative and non increasing.

There are some examples in the wiki where it happens to be useful
https://en.m.wikipedia.org/wiki/Cauchy_condensation_test

But it's a pretty unusual situation. The test is basically just a generalization of the normal way people prove that the harmonic series diverges.

In practice, the ratio test and the integral test are used to solve almost every problem.

 

[Mark44]

In practice, the ratio test and the integral test are used to solve almost every problem.

A couple other tests are useful to know: the limit comparison test and the n-th term test for divergence. The latter test is very easy to use to determine that a series diverges; that is, if $\lim_n \to \infty a_n \ne 0$, then you can conclude that the series diverges. However, if $\lim_n \to \infty a_n = 0$, students often conclude, mistakenly, that the series converges.