Show that the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}##converges

[chwala]

$$=\lim_{n→∞}\left[ \dfrac{n}{(n+1)}⋅\dfrac{\ln n}{(\ln(n+1)}\right]$$
$$=\lim_{n→∞}\left[ \dfrac{n}{(n+1)}\right]⋅\lim_{n→∞}\left[ \dfrac{\ln n}{(\ln(n+1)}\right]$$
$$ =\lim_{n→∞}\left[ \dfrac{n}{(n+1)}\right]⋅\lim_{n→∞}\left[ \dfrac{n+1}{n}\right]$$
$$ =\lim_{n→∞}\left[ \dfrac{1}{(1+1/n)}\right]⋅\lim_{n→∞}\left[\dfrac{1+1/n}{1}\right]=\left[\dfrac{1}{1+0}⋅\dfrac{1+0}{1}\right]=1⋅1=1$$
Bingo

 

[docnet]

That is what I used...though not indicated in working ...

On the $5$th line of post $26$, it looks like you used the product rule of differentiation yet somehow ended up with the correct limit in the end. From the $4$th line of post $26$,
$$\displaystyle{\lim_{n \to \infty}}[\dfrac{n}{(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1))}] =\displaystyle{\lim_{n \to \infty}}[\dfrac{\frac{d}{dn}(n)}{\frac{d}{dn}(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1))}]=1\cdot \displaystyle{\lim_{n \to \infty}}[\dfrac{\ln n}{(\ln(n+1))}]=\displaystyle{\lim_{n \to \infty}}[\dfrac{\frac{d}{dn}(\ln n)}{\frac{d}{dn}(\ln(n+1))}]$$
$$=\displaystyle{\lim_{n \to \infty}}\frac{n+1}{n} =\displaystyle{\lim_{n \to \infty}}\frac{\frac{d}{dn}(n+1)}{\frac{d}{dn}(n)} =\displaystyle{\lim_{n \to \infty}}\frac{1}{1}=1$$

 

[Mark44]

... $=\lim_{n→∞} [\dfrac{n}{(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1)}]$
$=\lim_{n→∞} [\dfrac{n}{(n+1)}]⋅\lim_{n→∞}[\dfrac{\ln n}{(\ln(n+1)}]$
$=\lim_{n→∞} [\dfrac{n}{(n+1)}]⋅\lim_{n→∞}[\dfrac{n+1}{n}]$
$=\lim_{n→∞} [\dfrac{1}{(1+1/n)}]⋅\lim_{n→∞}[\dfrac{1+1/n}{1}]=\dfrac{1}{1+0}⋅\dfrac{1+0}{1}=1⋅1=1$
Bingo

Quicker...
$\lim_{n→∞} [\dfrac{n}{(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1)}]$
$=\lim_{n→∞} [\dfrac{n}{n(1 +1/n)}]⋅\lim_{n→∞}[\dfrac{\ln n}{(\ln(n+1)}]$
$=1⋅\lim_{n→∞}[\dfrac{n+1}{n}]$ (L'Hopital's Rule used here)
$=1⋅\lim_{n→∞}[1+1/n]=1⋅1=1$

You can save a bit of writing by getting some of the calculations out of the way rather than dragging them along.

 

[chwala]

On the $5$th line of post $26$, it looks like you used the product rule of differentiation yet somehow ended up with the correct limit in the end. From the $4$th line of post $26$,
$$\displaystyle{\lim_{n \to \infty}}[\dfrac{n}{(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1))}] =\displaystyle{\lim_{n \to \infty}}[\dfrac{\frac{d}{dn}(n)}{\frac{d}{dn}(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1))}]=1\cdot \displaystyle{\lim_{n \to \infty}}[\dfrac{\ln n}{(\ln(n+1))}]=\displaystyle{\lim_{n \to \infty}}[\dfrac{\frac{d}{dn}(\ln n)}{\frac{d}{dn}(\ln(n+1))}]$$
$$=\displaystyle{\lim_{n \to \infty}}\frac{n+1}{n} =\displaystyle{\lim_{n \to \infty}}\frac{\frac{d}{dn}(n+1)}{\frac{d}{dn}(n)} =\displaystyle{\lim_{n \to \infty}}\frac{1}{1}=1$$

Yes, product rule it is...

 

[BvU]

... $=\lim_{n→∞} [\dfrac{n}{(n+1)}]⋅[\dfrac{\ln n}{(\ln(n+1)}]$
etc
Bingo

Hey hey hey, I want some more attention for the typesetting !

• use $$ instead of $to get$\displaystyle \lim_{n\rightarrow\infty}$instead of$\lim_{n\rightarrow\infty} \phantom{\Biggl | } ##
• use \left [ and \right ] to get $\left [\dfrac{n+1}{n}\right ]$ instead of $[\dfrac{n+1}{n}]$

$\$

 

[chwala]

Hey hey hey, I want some more attention for the typesetting !

• use $$ instead of $to get$\displaystyle \lim_{n\rightarrow\infty}$instead of$\lim_{n\rightarrow\infty} \phantom{\Biggl | } ##
• use \left [ and \right ] to get $\left [\dfrac{n+1}{n}\right ]$ instead of $[\dfrac{n+1}{n}]$

$\$

Thanks a lot Bvu, I've learned quite a lot from you in regards to Latex...I will for sure diarize all the syntax in my notebook...I will equally amend my posts accordingly...

 

[SammyS]

Thanks a lot BvU, I've learned quite a lot from you in regards to Latex...I will for sure diarize all the syntax in my notebook...I will equally amend my posts accordingly...

By the way, if you want to use LaTeX in the same line as normal text, but you want the look produced by $$, use the "\displaystyle" command. That is what @BvU used to get $\displaystyle \lim_{n \to \infty}$ to appear in a line of normal text.

You can also save some typing by using "\to" rather than "\rightarrow" for the arrow.

 

[chwala]

By the way, if you want to use LaTeX in the same line as normal text, but you want the look produced by $$, use the "\displaystyle" command. That is what @BvU used to get $\displaystyle \lim_{n \to \infty}$ to appear in a line of normal text.

You can also save some typing by using "\to" rather than "\rightarrow" for the arrow.

Am now a disciple of $$.

 

[SammyS]

Am now a disciple of $$.

I avoid $$ whenever possible.

 

[chwala]

I avoid $$ whenever possible.

Noted Sammy ...cheers

 

[Mark44]

I avoid $$ whenever possible.

Me, too. I'll use it if I have an equation I want to set off, but otherwise I will use ## so as to not incorporate a bunch of extraneous blank lines.
@SammyS, I didn't know that about displaystyle...

 

[anuttarasammyak]

Though It may be too late to join I would say
$$\sum_{n=1}\frac{1}{(n+1)^2}<\int_1^\infty \frac{1}{x^2}dx=1<\sum_{n=1}\frac{1}{n^2}$$
and
$$\sum_{n=1}\frac{1}{n^2}=1+\sum_{n=1}\frac{1}{(n+1)^2}$$
So
$$1<\sum_{n=1}\frac{1}{n^2}<2$$
, the series converges.

 

[chwala]

Bvu/Sammy can you kindly post me an example of a problem posted with;
"\displaystyle" command (two lines will suffice)...i do not seem to get how the command should be typed...
Alternatively, I will just peruse your posts and see how you are using that.. I am a fast learner. Cheers mate

 

[BvU]

By the way, if you want to use LaTeX in the same line as normal text, but you want the look produced by $$, use the "\displaystyle" command. That is what @BvU used to get $\displaystyle \lim_{n \to \infty}$ to appear in a line of normal text.

You can also save some typing by using "\to" rather than "\rightarrow" for the arrow.

Right-click on this \lim thing !

 

[BvU]

Though It may be too late to join I would say
...

Never too late ! And a nice warming-up for part (c) in this $\TeX$/series convergence thread !

 

[BvU]

I think we wwant @chwala to catch up first ...
It seems to me part (c) promises convergence ...

 

[chwala]

I think we wwant @chwala to catch up first ...
It seems to me part (c) promises convergence ...

Yes definitely, I'll be posting my findings soon...

 

[anuttarasammyak]

Similar to #42
$$\sum_{n=1}\frac{1}{(n+1)^p}<\int_1^\infty \frac{1}{x^p}dx=[\frac{x^{1-p}}{1-p}]_1^\infty<\sum_{n=1}\frac{1}{n^p}$$
and
$$\sum_{n=1}\frac{1}{n^p}=1+\sum_{n=1}\frac{1}{(n+1)^p}$$
for p>0. So the series converges for p>1, e.g. p=1.01.

 

[chwala]

Ok, for part $c$,
The integral test can be used because;
1. series is continous.
2. series is easily integrable by use of improper integrals.
We want to check the convergence or divergence of,
$$\sum_{n=2}^\infty\left[\dfrac{1}{n\ln(n)}\right]$$ Using the integral test it follows that,
$$\int_2^\infty\left[\dfrac{1}{n\ln(n)}\right] dn=\lim_{m→∞}\int_2^m\left[\dfrac{1}{x\ln(x)}\right] dx$$
Let$u= \ln(x)$ then $du$=$\dfrac{1}{x}dx$, therefore
$$\lim_{m→∞}\int_2^m\left[\dfrac{1}{x\ln(x)}\right] dx=\lim_{m→∞}\int_2^m\left[\dfrac{du}{u}\right]=\lim_{m→∞}\left[\ln(\ln(m)-\ln(\ln2)\right]=∞$$
The integral diverges, therefore the series diverges.

 

[anuttarasammyak]

Using the integral test it follows that,
∫2∞[1nln⁡(n)]dn=limm→∞∫2m[1xln⁡(x)]dx

In more detail
$$\sum_{n=2}^\infty \frac{1}{(n+1)\ln (n+1)}<\int_2^\infty\frac{1}{x \ln x}dx=[\ln(\ln x)]_2^\infty=+\infty< \sum_{n=2}^\infty \frac{1}{n\ln n}$$
and obviously
$$\sum_{n=2}^\infty \frac{1}{n\ln n}= \sum_{n=2}^\infty \frac{1}{(n+1)\ln (n+1)}+\frac{1}{2 \ln 2}$$
So the series diverges.

 

[BvU]

Hehe, so my playing advocate for the devil in #46 hasn't confused anyone but myself .
Well done @chwala and @anuttarasammyak !

$\$

 

[chwala]

Hehe, so my playing advocate for the devil in #46 hasn't confused anyone but myself .
Well done @chwala and @anuttarasammyak !

$\$

Nice Bvu...Great Math community that we have here...I also noted that we can integrate $\dfrac {1}{x\ln(x)}$ using integration by parts...and realize the same result.