Question

series convergence/divergence, Calculus...

NOTE(S):
Example #6, has a correction later on ... (please look there)

Answer 1

I will be adding some examples from time to time.
Will try to keep the thread clean as much as possible.

Answer 2

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}1 }}\)
\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{2n}{n^2+1}}\)

It already seems like the series is going to diverge,
the question is, \(\color{black}{\sf how~~to~~show~~this?}\)

We will use the "Comparison Test". Let's make a smaller series.
For example the one below, which as you see is easy to simplify.

\(\color{black}{ \displaystyle B_n=\sum_{ n=1 }^{ \infty } \frac{2n}{n^2+n^2}\quad \Rightarrow \quad \sum_{ n=1 }^{ \infty } \frac{2n}{2n^2}\quad \Rightarrow \quad \sum_{ n=1 }^{ \infty } \frac{1}{n}}\)

Since \(\color{black}{B_n}\) is smaller than \(\color{black}{A_n}\), and diverges,
therefore \(\color{black}{A_n}\) also diverges.

Answer 3

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}2 }}\)
\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{n^{40}}{e^n}}\)

\(\color{black}{\sf In~fact,~this~converges!}\) I will introduce a new
theory through this example.


However you know, \(\color{black}{ \displaystyle 400^{40}<2^{400}}\). \({\small\\[0.8em]}\)
And if, \(\color{black}{ \displaystyle \sum_{ n=400 }^{ \infty } \frac{n^{40}}{e^n}}\) converges, then \(\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{n^{40}}{e^n}}\) also converges. \({\small\\[0.8em]}\)
So we just have to show the convergence of \(\color{black}{ \displaystyle \sum_{ n=400 }^{ \infty } \frac{n^{40}}{e^n}}\).

Now we can use the comparison test.
\(\color{black}{ \displaystyle \sum_{ n=400 }^{ \infty } \frac{n^{40}}{e^n}\le \sum_{ n=400 }^{ \infty } \frac{2^n}{e^n}}\) \(\small\\[1.7em]\)
\(\color{black}{ \displaystyle \sum_{ n=400 }^{ \infty } \frac{2^n}{e^n}=(2/e)^n\Longrightarrow {\rm Converges}}\)
(Because \(|r|<1\))

And since \(\color{black}{ \displaystyle 2^n/e^n}\) converges,
\(\color{black}{ \displaystyle n^{40}/e^n}\) also converges.
(((Whether starting from n=1 or n=400)))

So, the answer is: \(\color{black}{ \displaystyle A_n}\) converges.

Answer 4

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}3 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{5n^3}{n^4-2n^3+n^2+1}}\)

You can tell the series is going to diverge,
because of its leading powers:

(It will be something like: \(\color{black}{ \displaystyle n^3/n^4=1/n}\)
- the "Harmonic Series" which we know Diverges)


So, by the Comparison Test we will compare \(\color{black}{ \displaystyle A_n}\)
to some smaller series \(\color{black}{ \displaystyle B_n}\).

\(\color{black}{ \displaystyle B_n=\sum_{ n=1 }^{ \infty } \frac{5n^3}{n^4+2n^4+n^4+n^4}=\sum_{ n=1 }^{ \infty } \frac{5n^3}{5n^4}}\)

\(\color{black}{ \displaystyle B_n=\sum_{ n=1 }^{ \infty } \frac{1}{n}}\)

\(\color{black}{ \displaystyle B_n}\) is smaller than \(\color{black}{ \displaystyle A_n}\), and \(\color{black}{ \displaystyle B_n}\) diverges,
THEREFORE, the answer is: \(\color{black}{ \displaystyle A_n}\) diverges.

Answer 5

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}4 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{n!}{{e^n}^{^2}}}\)

you will be surprised how quickly
I`m going to tackle this series ...\(\tiny{\\[2.1em]}\)
One posssible way to do this, is through the
Comparison and then the Nth Root Test.\(\tiny{\\[3.8em]}\)
I know that \(\color{black}{ \displaystyle n^n\ge n! }\),
So I am going to set, \(\color{black}{ \displaystyle B_n= \sum_{ n=1 }^{ \infty } \frac{n^n}{{e^n}^{^2}}}\)
(Note that \(\color{black}{ \displaystyle B_n\ge A_n}\) )
\(\tiny{\\[1.4em]}\)
So, by the Comparison Test if \(\color{black}{ \displaystyle B_n}\) converges,
then \(\color{black}{ \displaystyle A_n}\) also converges.\(\tiny{\\[1.9em]}\)
To show the convergence of \(\color{black}{ \displaystyle B_n}\), I will use
the Nth Root test.

\(\color{black}{ \displaystyle \lim_{n\to\infty} \left|~\sqrt[n]{B_n{\color{white}{\large \ell}}}~\right|=\lim_{n\to\infty}\left|~\sqrt[n]{\frac{n^n}{{e^n}^{^2}}}~\right|=\lim_{n\to\infty}\left|\frac{n}{e^n}\right|=0}\)

By L'Hospital's rule, this limit is zero.

\(\color{red}{((}\)Because as it is now, the limit is ∞/∞
and when I differentiate top and bottom
I get 1/e\(\rm ^n\); therefore this limit is =0\(\color{red}{))}\)

\(\color{black}{ \displaystyle B_n}\) converges \((\)by the Nth Root Test\()\), and
\(\color{black}{ \displaystyle B_n}\) is larger than \(\color{black}{ \displaystyle A_n}\). Therefore, \(\color{black}{ \displaystyle A_n}\) converges.

Answer 6

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}5 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{n^n}{(n+1)^{n+1}}}\)\(\tiny{\\[3.5em]}\)
This series might seem quite difficult, however
we will nail the series with Comparison Test.\(\tiny{\\[2.0em]}\)
\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{n^n}{(n+1)^{n+1}} \quad \Rightarrow \quad A_n=\sum_{ n=1 }^{ \infty }\left[\frac{1}{n+1}\times \frac{n^n}{(n+1)^{n}}\right]}\)
\(\tiny{\\[1.4em]}\)
You know that, (for all n\(\ge\)0)\(\tiny{\\[1.3em]}\)
\(\color{black}{ \displaystyle 1>\frac{n^n}{~~(n+1)^{n~}}>\frac{1}{e} }\)

So, we can make the following \(\color{black}{ \displaystyle B_n}\) which is
smaller (than \(\color{black}{ \displaystyle A_n}\)), and yet this \(\color{black}{ \displaystyle B_n}\) will diverge
(and this will show the divergence of \(\color{black}{ \displaystyle A_n}\))

\(\color{black}{ \displaystyle B_n=\sum_{ n=1 }^{ \infty }\left[\frac{1}{n+1}\times \frac{1}{e}\right]}\)\(\tiny{\\[3.9em]}\)
\(\color{black}{ \displaystyle B_n=\frac{1}{e}\sum_{ n=2 }^{ \infty }\frac{1}{n} }\) (and this clearly diverges)

\(\color{black}{ \displaystyle B_n}\) is smaller than \(\color{black}{ \displaystyle A_n}\), and \(\color{black}{ \displaystyle B_n}\) diverges.
Therefore, \(\color{black}{ \displaystyle A_n}\) also diverges.

Answer 7

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}6 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=2 }^{ \infty } \frac{\cos(n)}{\sqrt{~~n{\color{white}{\large >}}}} }\)

You can tell the series is going to converge,
b/c this series is kind of an alternating series.
((something like non-integer alternation,
although there isn't such a term officially, but
I hope you know what I mean, in any case ...))

In any case, we can find the integral of \(\color{black}{ \displaystyle A_n}\)
and if it converges then \(\color{black}{ \displaystyle A_n}\) will also converge.

\(\color{black}{ \displaystyle {\cos(n)}=\sum_{k=0}^{\infty}\frac{(-1)^k\cdot n^{2k}}{(2k)!} }\) \(\tiny{\\[3.9em]}\)
\(\color{black}{ \displaystyle {n^{-1/2}\cos(n)}=\sum_{k=0}^{\infty}\frac{(-1)^k\cdot n^{2k-1/2}}{(2k)!} }\) \(\tiny{\\[3.9em]}\)
\(\color{black}{ \displaystyle \int_0^\infty {n^{-1/2}\cos(n)}{\rm ~ d}n=\sum_{k=0}^{\infty}\frac{(-1)^k\cdot n^{2k+1/2}}{(2k)!\times (2k+1/2)} }\)

And now, I will apply the Comparison Test to
show why the integral of \(\color{black}{ \displaystyle A_n}\), (which I wrote as
a series), will converge. ((And recall that I need
to show this convergence in order to conclude
that \(\color{black}{ \displaystyle A_n}\) converges by the Integral Test))
\(\tiny{\\[1.3em]}\)
\(\color{black}{ \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^k\cdot n^{2k+1/2}}{(2k)!\times (2k+1/2)}\ge \sum_{k=0}^{\infty}\frac{n^{3k}}{k!}=e^{\large\left\{n^3\right\} } }\)

The very last answer: \(\color{black}{ \displaystyle e^{\large\left\{n^3\right\} } }\)\(,\) is based on the
following well known series.\(\tiny{\\[1.6em]}\)
\(\large\color{black}{ \displaystyle \sum_{k=0}^{\infty}\frac{y^{k}}{k!}=e^{\large\left\{y\right\} } }\) ((and in that case \(y=n^3\)))

So, LET`S go over everything and CONCLUDE.

Our original series is, \(\color{black}{ \displaystyle A_n= \sum_{ n=2 }^{ \infty } \frac{\cos(n)}{\sqrt{~~n{\color{white}{\large >}}}} }\)

I have showed \(\color{black}{ \displaystyle A_n}\) converges by the Integral Test.
And the integral of \(\color{black}{ \displaystyle A_n}\) was \(\tiny\\[1.5em]\)
\(\color{black}{ \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^k\cdot n^{2k+1/2}}{(2k)!\times (2k+1/2)} }\)

We have showed that this integral-series converges,
by comparing that to a new (bigger) series, which was \(\tiny{\\[1.5em]}\)
\(\color{black}{ \displaystyle \sum_{k=0}^{\infty}\frac{n^{3k}}{k!} }\) ((which converged to \(e^{\large\left\{n^3\right\}}\)))

So, THE CONCLUSION is:
Since the integral of \(\color{black}{ \displaystyle A_n}\) converges (by the
Comparison Test), therefore \(\color{black}{ \displaystyle A_n}\) (itself) converges.

Answer 8

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}7 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=2}^{ \infty } \frac{1}{\ln(n!)}}\) \(\tiny{\\[3.5em]}\)
\(\color{black}{ \displaystyle n!<n^n\quad \Longrightarrow \quad \frac{1}{n!}>\frac{1}{n^n} }\) \(\tiny{\\[3.5em]}\)
So will let, \(\color{black}{ \displaystyle B_n= \sum_{ n=2}^{ \infty } \frac{1}{\ln(n^n)}=\sum_{ n=2}^{ \infty } \frac{1}{n\ln(n)}}\) \(\tiny{\\[3.7em]}\)
(note that \(\color{black}{ \displaystyle B_n<A_n {~~} }\) for all n>2) \(\tiny{\\[2.0em]}\)
When you integrate \(\color{black}{ \displaystyle \int_{ n=2}^{n=\infty } \frac{1}{n\ln(n)}{\rm d}n}\) \(\tiny{ \\[3.0em] }\)
(set \(u=\ln n\)), and you will get that: \(\tiny{\\[2.6em]}\)
\(\color{black}{ \displaystyle \int_{ n=2}^{n=\infty } \left.\frac{1}{n\ln(n)}=\ln \left\{\ln n \right\}\right|_{ n=2}^{n=\infty }\quad \Longrightarrow {\bf Diverges} }\)


Since \(\color{black}{ \displaystyle B_n <A_n}\), and \(\color{black}{ \displaystyle B_n }\) diverges (by the Integral Test),
therefore \(\color{black}{ \displaystyle A_n }\) diverges.

Answer 9

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}8 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=1 }^{ \infty } \frac{\ln\left(n^4{\tiny~}!\right)}{n^6}}\)

We will use \(\color{black}{ \displaystyle y(x)!<y(x)^{y(x)}}\) (for all x>1)
\(\bf (\)This is true for any function that increases.
It may be \(y=x^p;{\tiny~~}p>0\) or, \(y=a^x;{\tiny~~}a>1\)
or any other "increasing" function\(\bf )\)

Here, \(x\) is \(n^4\) and the statement is still true.
So, let's apply this....

\(\color{black}{ \displaystyle \frac{\ln\left(n^4{\tiny~}!\right)}{n^6}<\frac{\ln\left((n^4)^{^{\large \left(n^4\right)}}\right)}{n^6}=\frac{n^4\ln(n^4)}{n^6}=\frac{4n^4\ln(n)}{n^6} }\)

So, it comes out that ((\(n^4\) cancels))
\(\color{black}{ \displaystyle \frac{\ln\left(n^4{\tiny~}!\right)}{n^6}<\frac{4\ln(n)}{n^2} }\)

And for this reason, we will set
\(\color{black}{ \displaystyle B_n=\sum_{ n=1 }^{ \infty } \frac{4\ln(n)}{n^2} }\)

It is easy enough to show that:
\(\color{black}{ \displaystyle \lim_{ n\to\infty } \frac{\ln(n)}{n^p}=0}\) (provided that p>0)

You can show this yourself ....
\((1)\) Use L'Hospital's rule - (since it's ∞/∞)
\((2)\) Multiply top and bottom times \(n^1\)
\((3)\) Then, the answer is obvious

Therefore, \(\sqrt{n}>\ln(n)\), and we will use the
Comparison Test once again.
\(\color{black}{ \displaystyle C_n=\sum_{ n=1 }^{ \infty } \frac{4\sqrt{n}}{n^2} =\sum_{ n=1 }^{ \infty } \frac{4}{n^{1.5}} }\)

(and that is a convergent p-series)

So, let's conclude now.
Recall that,

\(\color{black}{ \displaystyle {\tiny~~~~~~~~}\color{#0033cc}{A_n}{\tiny~~~~~~~~}<{\tiny~~~~~~~~~}\color{#009933}{B_n}{\tiny~~~~~~}<{\tiny~~~~~~~}\color{#997300}{C_n} }\)

\(\color{black}{ \displaystyle \color{#0033cc}{\frac{\ln\left(n^4{\tiny~}!\right)}{n^6}}<\color{#009933}{\frac{4\ln(n)}{n^2}} <\color{#997300}{\frac{4}{n^{1.5}}} }\)

And since \(C_n\) converges (p-series; p>1)
therefore, ((\(B_n\) converges and)) \(A_n\) converges.

Answer 10

Can someone please remove that?

Answer 11

\(\Large \color{blue}{\bf From~this~points,~I~will~get~more~formal!}\)

Answer 12

\({\bbox[3pt, lightcyan ,border:2px solid black ]{ \rm Example~~\text{#}9 }}\)

\(\color{black}{ \displaystyle A_n= \sum_{ n=\ell}^{ \infty } \frac{1}{\ln(nx{\tiny~}!)}}\) \(\tiny{\\[3.5em]}\)
\(\color{black}{ \displaystyle nx{\tiny~}!<(nx)^{nx}\quad \Longrightarrow \quad \frac{1}{nx{\tiny~}!}>\frac{1}{(nx)^{nx}} }\) \(\tiny{\\[3.5em]}\)
So will let, \(\color{black}{ \displaystyle B_n= \sum_{ n=\ell}^{ \infty } \frac{1}{\ln\left[(nx)^{nx}\right]}=\sum_{ n=2}^{ \infty } \frac{1}{nx\ln(nx)}}\) \(\tiny{\\[3.7em]}\)
\(\color{black}{ \displaystyle nx\ln(nx)<nx\ln(n^2)=2xn\ln(n)\quad \Longrightarrow \quad \frac{1}{nx\ln(nx)}>\frac{1}{2xn\ln(n)} }\) \(\tiny{\\[3.5em]}\)
So will let, \(\color{black}{ \displaystyle C_n= \sum_{ n=\ell}^{ \infty } \frac{1}{2xn\ln(n)}}\) \(\tiny{\\[3.7em]}\)

When you integrate \(\color{black}{ \displaystyle \int_{ n=2}^{n=\infty } \frac{1}{2xn\ln(n)}{\rm d}n}\) \(\tiny{ \\[3.0em] }\)
(set \(u=\ln n\)), and you will get that: \(\tiny{\\[2.6em]}\)
\(\color{black}{ \displaystyle \int_{ n=2}^{n=\infty } \left.\frac{1}{2xn\ln(n)}=2x\ln \left\{\ln n \right\}\right|_{ n=\ell}^{n=\infty }\quad \Longrightarrow {\bf Diverges} }\)

Since \(\color{black}{ \displaystyle C_n<B_n <A_n}\), and \(\color{black}{ \displaystyle C_n }\) diverges (by the Integral Test),
therefore \(\color{black}{ \displaystyle A_n }\) diverges.

Answer 13

I don't think the proof for \(\displaystyle A_n= \sum_{ n=2 }^{ \infty } \frac{\cos(n)}{\sqrt{n}}\) converges is correct. The problem is that \(n^{-1/2}\cos(n)\) is alternating and integral test cannot be applied to an alternating series. Look at the graph attached. The blue curve is \(n^{-1/2}\cos(n)\). The integral (blue area under curve) does not envelope the whole series (red rectangles). Sometimes the red rectangles are larger than the blue area and sometime the blue area is larger than the red rectangles.

FYI, I tried root test and it doesn't work either.

Answer 14

\(\color{#ff0000 }{ \displaystyle \bf \LARGE Attention,~please ! }\)
\(\color{#ff0000 }{ \displaystyle \bf \LARGE Example~\text{#}6,~corrected. }\)
\(\color{#ff0000 }{ \displaystyle \bf \LARGE \left(Still~converges,~but....\right) }\)

Yes, I was indeed incorrect in solving \(\color{#000000 }{ \displaystyle \rm Example~\text{#}6 }\);
thank you for correcting me @thomas5267 , because
I applied the Integral Test to an alternating series,
(but, actually that isn't allowed).

\(\color{#003366 }{ \displaystyle \bf \LARGE Below~is~the~correct }\)
\(\color{#003366}{ \displaystyle \bf \LARGE solution~to~example~\text{#}6 }\)
\(\tiny \\[0.6 em]\)
\(\color{#000000 }{ \displaystyle A_n=\sum_{n=2}^{\infty}\frac{\cos(n)}{\sqrt{n}} }\)
\(\tiny \\[0.6 em]\)
You can try to take the partial sums of the series, and
it will converge, but to actually prove the convergence,
\(\color{#660033 }{\rm Dirichlet’s{\tiny~~~}Test}\) is used.
\(\tiny \\[1.0em]\)
In our case,\(\tiny \\[0.6 em]\)
\(\color{#000000 }{ \displaystyle x_n=\cos(n) }\) \(\tiny \\[0.6 em]\)
\(\color{#000000 }{ \displaystyle y_n=n^{-1/2} }\)

And these perfectly satisfy the requierements
of the Dirichlet's Test; meaning that,

At first, the sum of \(\color{#000000 }{ \displaystyle x_n }\) is (always) bound: \(\tiny \\[0.5 em] \)
\(\color{#000000 }{ \displaystyle \left|\sum_{n=2}^{\rm K}\cos(n) \right|\le 2 }\) this is true for all [natural number] \(\color{#000000 }{ \displaystyle {\rm K} }\).
\(\tiny \\[0.8 em] \)
At second, \(\color{#000000 }{ \displaystyle y_n }\) is always positive (since we are only
considering natural-number n), and \(\color{#000000 }{ \displaystyle y_n }\) satisfies,\(\tiny \\[0.5 em] \)
\(\color{#000000 }{ \displaystyle y_{n-1}>y_{n}~~\forall n }\) \(\tiny \\[0.6 em] \)
\(\color{#000000 }{ \displaystyle \lim_{n\to\infty} y_{n}=0 }\)

So, the series is convergent.

Answer 15

How do you derive bound of \(\displaystyle\left|\sum_{n=2}^K\cos(n)\leq2\right|\)?

Answer 16

Don't want to devote to much space to this, (rather, I applied some logic,) however I bet that the actual bounds might be only a little larger.

THis is the [formal] derivation,
or so I would refer to it...

\(\color{#000000 }{ \displaystyle \sum_{n=0}^{\large \ell}\cos(nx)=\frac{1}{2}\left(\frac{\sin((\ell+{\large ½})x)-\sin(x/2)}{\sin (x/2)}\right) }\)

\(\color{#000000 }{ \displaystyle \sum_{n=0}^{\large \ell}\cos(n)=\frac{1}{2}\left[\frac{\sin(\ell+{\large ½})-\sin({\large ½})}{\sin ({\large ½})}\right] }\) (when x=1)

\(\color{#000000 }{ \displaystyle \sin(\ell+{\large ½}) \le 1 }\)

\(\color{#000000 }{ \displaystyle -\frac{1}{2}\left[\frac{1+\sin({\large ½})}{\sin ({\large ½})}\right] \le\sum_{n=0}^{\large \ell}\cos(n)\le\frac{1}{2}\left[\frac{1-\sin({\large ½})}{\sin ({\large ½})}\right] }\)

\(\color{#000000 }{ \displaystyle -1.54 \le\sum_{n=0}^{\large \ell}\cos(n)\le 0.54 }\)

And just to secure myself,

\(\color{#000000 }{ \displaystyle -2 \le\sum_{n=0}^{\large \ell}\cos(n)\le 2\quad \quad \Longrightarrow \quad \left|\sum_{n=0}^{\large \ell}\cos(n)\right|\le 2 }\)

\(\LARGE \color{#ff0000}{☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼}\)
Another method of deriving the bounds you can possibly consider if you just don't want to waste paper and time is, to use the fact that \(\color{#000000 }{ \displaystyle -1\le \cos(n) \le 1 }\).

Only 3 numbers in a row will maintain the same sign.
So, for any integer \(\ell\),
\(\color{#000000 }{ \displaystyle \left|\cos(\ell)+\cos(\ell+1) +\cos(\ell+2)\right| \le 3 }\)
And the next three terms will have the opposite sign, so
\(\color{#000000 }{ \displaystyle \left|\cos(\ell)+{\bf ...}+\cos(\ell+5)\right| < 1 }\)
(Would be common sense for a 'periodic' function with bound area)

So you can tell that

\(\color{#000000 }{ \displaystyle \left|\sum_{n=0}^{\large \ell}\cos(n)\right|\le 3 }\)
\(\LARGE \color{#ff0000}{☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼}\)

(I don't think the calculus professor would approciate such impudence)