golgiapparatus31
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Show that sum from 1 to infinity of [ ln n / (n+4) ] diverges.

I try the integral test but I'm not able to integrate by parts.. How do I show this? Please help
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RDKGames
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(Original post by golgiapparatus31)
Show that sum from 1 to infinity of [ln n / (n+4)] diverges.

I try the integral test but I'm not able to integrate by parts.. How do I show this? Please help
\begin{aligned} \sum_{n=1}^{\infty} \dfrac{\ln n}{n+4} & > \sum_{n=5}^{\infty} \dfrac{\ln n}{n+4} \\ & > \sum_{n=5}^{\infty} \dfrac{\ln n}{n+n} > 0 \end{aligned}

This last sum diverges due to integral test.

Note that \displaystyle \int \dfrac{\ln x}{x} \ dx =  \dfrac{1}{2} (\ln x)^2 + c
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golgiapparatus31
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(Original post by RDKGames)
\begin{aligned} \sum_{n=1}^{\infty} \dfrac{\ln n}{n+4} & > \sum_{n=5}^{\infty} \dfrac{\ln n}{n+4} \\ & > \sum_{n=5}^{\infty} \dfrac{\ln n}{n+n} > 0 \end{aligned}

This last sum diverges due to integral test.

Note that \displaystyle \int \dfrac{\ln x}{x} \ dx =  \dfrac{1}{2} (\ln x)^2 + c
Thank you!! (PRSOM)

Can you please check my answer to this too?

Using the comparison test, or otherwise, show that the sum from 1 to infinity of (1/(sqrt(n2+3) diverges.

1/sqrt(n2+3) > 1/sqrt(n2+4)
Integrating 1/sqrt(x2+4) gives sinh-1(x/2) so diverges. Hence the sum diverges


Evaluate whether the sum from 1 to infinity of (3n / (2n + 5n)) diverges or converges.

I start by writing that
(3x / (2x + 5x)) < 3x/5x = (3/5)x
because making the denominator smaller makes the value of the fraction larger

Then I calculate the integral of 0.6^x from 1 to infinity, getting 1.17. So the sum converges.

Does this look correct? Thank you!!
I integrate (3/5)^x from 1 to infinity,
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RDKGames
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(Original post by golgiapparatus31)
Thank you!! (PRSOM)

Can you please check my answer to this too?

Using the comparison test, or otherwise, show that the sum from 1 to infinity of (1/(sqrt(n2+3) diverges.

1/sqrt(n2+3) > 1/sqrt(n2+4)
Integrating 1/sqrt(x2+4) gives sinh-1(x/2) so diverges. Hence the sum diverges
Regarding \displaystyle \sum_{n\in\mathbb{N}} \dfrac{1}{\sqrt{n^2 + 3}}, your proof looks good.

Though why didn't you just immediately integrate into \mathrm{arcsinh} \ \dfrac{x}{\sqrt{3}} ?

Personally, I would've just used a comparison test.

\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n^2 + 3}} &gt; \sum_{n = 2}^{\infty} \dfrac{1}{\sqrt{n^2 + 3}} &gt; \sum_{n = 2}^{\infty} \dfrac{1}{\sqrt{n^2 + n^2}} = \dfrac{1}{\sqrt{2}} \sum_{n = 2}^{\infty} \dfrac{1}{n}

The harmonic series is well known to diverge. Part of me wants to say this approach is a bit more natural for a real analysis course because inverse sinh may not be an 'obvious' or a studied function whose properties you can just read off without proof.
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DFranklin
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(Original post by golgiapparatus31)
Thank you!! (PRSOM)

Can you please check my answer to this too?

Using the comparison test, or otherwise, show that the sum from 1 to infinity of (1/(sqrt(n2+3) diverges.

1/sqrt(n2+3) > 1/sqrt(n2+4)
Integrating 1/sqrt(x2+4) gives sinh-1(x/2) so diverges. Hence the sum diverges


Evaluate whether the sum from 1 to infinity of (3n / (2n + 5n)) diverges or converges.

I start by writing that
(3x / (2x + 5x)) < 3x/5x = (3/5)x
because making the denominator smaller makes the value of the fraction larger

Then I calculate the integral of 0.6^x from 1 to infinity, getting 1.17. So the sum converges.

Does this look correct? Thank you!!
I integrate (3/5)^x from 1 to infinity,
So, to some extent this is a "style" complaint, but since an integral is actually a pretty complex mathematical "thing" that's defined in terms of limits of sums, it's better style to use direct comparison where possible.

For these particular sums:

\dfrac{\ln x}{n+4} &gt; \dfrac{1}{2n} for n > 4, \dfrac{1}{\sqrt{n^2+4}} &gt; \dfrac{1}{3n} for n > 1 (so both diverge due to comparison with 1/n), while for your last case you get convergence since you're comparing to a GP with |ratio| < 1.

Even if you think I'm nitpicking about "style" (what you've done isn't actually wrong), these manipulations/comparisions are usually less work than the integral test and also help you with the kind of approach you'll use increasingly as you study analysis. (A large part of real analysis is finding "nice" functions that bound "not-so-nice" functions).
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RDKGames
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(Original post by golgiapparatus31)
Evaluate whether the sum from 1 to infinity of (3n / (2n + 5n)) diverges or converges.

I start by writing that
(3x / (2x + 5x)) < 3x/5x = (3/5)x
because making the denominator smaller makes the value of the fraction larger

Then I calculate the integral of 0.6^x from 1 to infinity, getting 1.17. So the sum converges.

Does this look correct? Thank you!!
I integrate (3/5)^x from 1 to infinity,
Regarding \displaystyle \sum_{n = 1}^{\infty} \dfrac{3^n}{2^n + 5^n} you are correct to note an important bound;

\displaystyle \sum_{n = 1}^{\infty} \dfrac{3^n}{2^n + 5^n} &lt; \sum_{n = 1}^{\infty} \dfrac{3^n}{5^n}

But there is no need to apply the integral test here! The RHS is a geometric series ... we can be more simple than the integral test here in order to answer this question.
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golgiapparatus31
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(Original post by RDKGames)
Regarding \displaystyle \sum_{n\in\mathbb{N}} \dfrac{1}{\sqrt{n^2 + 3}}, your proof looks good.

Though why didn't you just immediately integrate into \mathrm{arcsinh} \ \dfrac{x}{\sqrt{3}} ?

Personally, I would've just used a comparison test.

\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n^2 + 3}} &gt; \sum_{n = 2}^{\infty} \dfrac{1}{\sqrt{n^2 + 3}} &gt; \sum_{n = 2}^{\infty} \dfrac{1}{\sqrt{n^2 + n^2}} = \dfrac{1}{\sqrt{2}} \sum_{n = 2}^{\infty} \dfrac{1}{n}

The harmonic series is well known to diverge. Part of me wants to say this approach is a bit more natural for a real analysis course because inverse sinh may not be an 'obvious' or a studied function whose properties you can just read off without proof.
Thank you! In my book an example was worked out in that way: https://imgur.com/a/pHjZkpg and https://imgur.com/a/eyDuONc, so I first stated a function less than 1/sqrt(n^2+3), then I integrate it.

My book says: "When an alternative expression is used to evaluate a sum, it is known as a comparison test."

Since question asked for comparison test, I did it that way.. :s The book didn't do manipulation like you did so I didn't know that approach.

(Original post by RDKGames)
Regarding \displaystyle \sum_{n = 1}^{\infty} \dfrac{3^n}{2^n + 5^n} you are correct to note an important bound;

\displaystyle \sum_{n = 1}^{\infty} \dfrac{3^n}{2^n + 5^n} &lt; \sum_{n = 1}^{\infty} \dfrac{3^n}{5^n}

But there is no need to apply the integral test here! The RHS is a geometric series ... we can be more simple than the integral test here in order to answer this question.
Thank you!

(Original post by DFranklin)
So, to some extent this is a "style" complaint, but since an integral is actually a pretty complex mathematical "thing" that's defined in terms of limits of sums, it's better style to use direct comparison where possible.

For these particular sums:

\dfrac{\ln x}{n+4} &gt; \dfrac{1}{2n} for n > 4, \dfrac{1}{\sqrt{n^2+4}} &gt; \dfrac{1}{3n} for n > 1 (so both diverge due to comparison with 1/n), while for your last case you get convergence since you're comparing to a GP with |ratio| < 1.

Even if you think I'm nitpicking about "style" (what you've done isn't actually wrong), these manipulations/comparisions are usually less work than the integral test and also help you with the kind of approach you'll use increasingly as you study analysis. (A large part of real analysis is finding "nice" functions that bound "not-so-nice" functions).
Thank you! I am not sure how you reached this inequality:

"\dfrac{\ln n}{n+4} &gt; \dfrac{1}{2n} for n > 4, \dfrac{1}{\sqrt{n^2+4}} &gt; \dfrac{1}{3n} for n > 1"

How did you get that \dfrac{\ln n}{n+4} &gt; \dfrac{1}{2n} ?
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RDKGames
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(Original post by golgiapparatus31)
Thank you! I am not sure how you reached this inequality:

"\dfrac{\ln n}{n+4} &gt; \dfrac{1}{2n} for n > 4, \dfrac{1}{\sqrt{n^2+4}} &gt; \dfrac{1}{3n} for n > 1"

How did you get that \dfrac{\ln n}{n+4} &gt; \dfrac{1}{2n} ?
When n &gt; 4, we have:

\ln n &gt; \ln 4 &gt; \ln e = 1

n + 4 &lt; n + n = 2n \implies \dfrac{1}{n+4} &gt; \dfrac{1}{2n}

Multiplying these inequalities gives the inequality he stated.
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golgiapparatus31
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Thanks a lot for the insights you gave me, RDKGames asnd DFranklin!!

I will be reading more on this topic to improve my understanding


Thank you very much !!
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