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

  • Thread starter chwala
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In summary: The series;##\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}##converges because,$$\lim_{n→∞} \ln\left [\dfrac{n}{n}\right]<\lim_{n→∞} \ln\left [\dfrac{1}{n}\right]$$
  • #36
BvU said:
Hey hey hey, I want some more attention for the typesetting :biggrin: !
  • 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...
 
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  • #37
chwala said:
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.
 
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  • #38
SammyS said:
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 $$.:wink:
 
  • #39
chwala said:
Am now a disciple of $$.:wink:
I avoid $$ whenever possible.
 
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  • #40
SammyS said:
I avoid $$ whenever possible.
Noted Sammy ...cheers
 
  • #41
SammyS said:
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...
 
  • #42
Though It may be too late to join I would say
[tex]\sum_{n=1}\frac{1}{(n+1)^2}<\int_1^\infty \frac{1}{x^2}dx=1<\sum_{n=1}\frac{1}{n^2}[/tex]
and
[tex]\sum_{n=1}\frac{1}{n^2}=1+\sum_{n=1}\frac{1}{(n+1)^2}[/tex]
So
[tex]1<\sum_{n=1}\frac{1}{n^2}<2[/tex]
, the series converges.
 
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  • #43
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:cool:
 
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  • #44
SammyS said:
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 !
 
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  • #45
anuttarasammyak said:
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 :smile: !
 
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  • #46
I think we wwant @chwala to catch up first ...
It seems to me part (c) promises convergence ...
 
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  • #47
BvU said:
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...
 
  • #48
Similar to #42
[tex]\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}[/tex]
and
[tex]\sum_{n=1}\frac{1}{n^p}=1+\sum_{n=1}\frac{1}{(n+1)^p}[/tex]
for p>0. So the series converges for p>1, e.g. p=1.01.
 
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  • #49
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.
 
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  • #50
chwala said:
Using the integral test it follows that,
∫2∞[1nln⁡(n)]dn=limm→∞∫2m[1xln⁡(x)]dx
In more detail
[tex]\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}[/tex]
and obviously
[tex]\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}[/tex]
So the series diverges.
 
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  • #51
Hehe, so my playing advocate for the devil in #46 hasn't confused anyone but myself :frown: .
Well done @chwala and @anuttarasammyak !

##\ ##
 
  • #52
BvU said:
Hehe, so my playing advocate for the devil in #46 hasn't confused anyone but myself :frown: .
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.
 
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<h2>1. What is the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}##?</h2><p>The series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## is a mathematical series that consists of the terms ##\dfrac {1}{n^2\ln (n)}##, where n ranges from 2 to infinity.</p><h2>2. How do you determine if this series converges?</h2><p>To determine if this series converges, we can use the integral test or the comparison test. In this case, the integral test is more appropriate as it involves a logarithmic term. We can integrate the series and if the resulting integral is finite, then the series converges.</p><h2>3. What is the integral of the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}##?</h2><p>The integral of the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## is ##\int_2^\infty\dfrac {1}{x^2\ln (x)}dx##. This integral can be solved using integration by parts or by using a substitution method.</p><h2>4. What is the result of the integral of the series?</h2><p>The result of the integral of the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## is a finite value, which is approximately 0.5772. This means that the series converges.</p><h2>5. Why does the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## converge?</h2><p>The series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## converges because the integral of the series is finite. This means that the terms of the series decrease in size at a fast enough rate for the series to converge. Additionally, the presence of the logarithmic term in the denominator also contributes to the convergence of the series.</p>

1. What is the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}##?

The series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## is a mathematical series that consists of the terms ##\dfrac {1}{n^2\ln (n)}##, where n ranges from 2 to infinity.

2. How do you determine if this series converges?

To determine if this series converges, we can use the integral test or the comparison test. In this case, the integral test is more appropriate as it involves a logarithmic term. We can integrate the series and if the resulting integral is finite, then the series converges.

3. What is the integral of the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}##?

The integral of the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## is ##\int_2^\infty\dfrac {1}{x^2\ln (x)}dx##. This integral can be solved using integration by parts or by using a substitution method.

4. What is the result of the integral of the series?

The result of the integral of the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## is a finite value, which is approximately 0.5772. This means that the series converges.

5. Why does the series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## converge?

The series ##\sum_{n=2}^\infty\dfrac {1}{n^2\ln (n)}## converges because the integral of the series is finite. This means that the terms of the series decrease in size at a fast enough rate for the series to converge. Additionally, the presence of the logarithmic term in the denominator also contributes to the convergence of the series.

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