hints

[vast citrus]

Sum of this series. Can I apply beta gamma?

[hushed tinsel]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[golden pilot]

$\sum_{k=0}^{\infty}\frac{4k+1}{(2k+1)(2k+2)(2k+3)}$ right

[jovial nebula]

Graph = G for Garbage

[golden pilot]

I'm thinking partial fraction decomposition

yeah you can decompose it

ill leave it to you now @vast citrus

[vast citrus]

Lol

[vast citrus]

Bro next steps are tricky

Taking LCM and all

[vast citrus]

@junior acorn

[junior acorn]

what did you figure out so far

[vast citrus]

A/(2k+1)+B/(2k+2)+c/(2k+3)

@junior acorn

[junior acorn]

have you solved it

[vast citrus]

Nope I can't

Is there any other method like gamma beta function?

[junior acorn]

are you just using words

[vast citrus]

No

I solved a similiar question

[junior acorn]

it's basic linear algebra, you've done this a hundred times already

[vast citrus]

Nope never

Let me tell you why i am having difficulties here

A(2k+2)(2k+3)+B(2k+1)(2k+3)+C(2k+1)(2k+2)=4k+1

i cannot solve it in varibles

[junior acorn]

3 variables, 3 equations

[vast citrus]

That's it if you want me to put random numbers then i will skip

[junior acorn]

this is uniquely soluble

[heady bluff]

A quick method is multiplying by 2k+1 on each side and evaluate for k=-1/2

You will get A

[vast citrus]

Where are you seeing 3 equation

[junior acorn]

= 0k^2 + 4k + 1

ring a bell?

[vast citrus]

Hmm a little bit

[junior acorn]

when are two polynomials equal?

[vast citrus]

Now i can compare

[junior acorn]

good

[vast citrus]

😂😂😂

How stupid i am

[junior acorn]

when you are comfortable with the technique you can also apply Rotor's isolation method

[heady bluff]

Or if we are being rigorous consider the rational fraction F=4X+1/(2X+1)(2X+1)(2X+3)=A/2X+1 +B/2X+2+C/2X+3 if you multiply by 2X+1 on each side and evaluate for X=-1/2 you have A and do the same for B, multiply by 2X+2 and evaluate for X=-1 etc.

It’s faster imo if you know the trick

[junior acorn]

a lot faster even

very useful trick when there are lots of terms

[junior acorn]

also you should explain why this series converges in the first place

but that's not hard

[vast citrus]

It converges if it gets a certain number

[junior acorn]

you will assume convergence along the way

[vast citrus]

But why?

It is easy

[vast citrus]

(2k+1) multiplying

I don't understand why x=-1/2

[junior acorn]

what variable do you want to isolate

plug in k=-1/2 here

some things disappear

what do you see?

[vast citrus]

Wow

Awesome

Rotor

[heady bluff]

Because for X=-1/2, 2X+1=0, if we multiply on each side by 2X+1 then when you evaluate at X=-1/2 the B and C constants vanish leaving only A=something

[vast citrus]

I liked your method

A=-1/4

I got

And it is easy

I will try now al method

[junior acorn]

i didn't get this

Ax1x2 = -1 implies A = -1/2

if you understand how to apply Rotor's suggestion, then you don't need to worry about it anymore

the problem is not about linear algebra, but series manipulation

$$ \frac{4k+1}{(2k+1)(2k+2)(2k+3)} = \frac{-1}{2(2k+1)} + \frac{3}{2k+2} + \frac{-5}{2(2k+3)} $$

[jovial nebula]

aL

[junior acorn]

this is how you can write the general term

$$ -\frac{1}{2}\sum _{k=0}^n \left( \frac{1}{2k+1} - \frac{6}{2k+2} + \frac{5}{2k+3}\right) $$

can you figure out what this is for all n?

[jovial nebula]

aL

[junior acorn]

@vast citrus

many paths to victory here, but I'll leave you with this for now

i got this answer, eventually

[vast citrus]

Bro

I don't want answers

😁😁

I can think of log series

[junior acorn]

says you who are obsessing over fake solves and whatnot 😂

[vast citrus]

If you want me to do a fakesolve i will add a few start terms and check with options....lol

[heady bluff]

How did you solve?

[vast citrus]

Just imaging all options in log terms so

log(1+x)=x-x^2/2+x^3/3-..

loh(1+x)/log(1-x)

I found it

Yooo

log(2/3)

[heady bluff]

Oh well I don’t know how you did but i see two ways of doing one is too express everything inside in terms of an integral then using the intervertion of series integral theorem to intervert both then get the result

[junior acorn]

that's probably what I did but I don't recognise this phrase

[heady bluff]

Second method is a bit similar, we can rearrange the terms in the sum to get a telescopic sum and a sum of something that can be rewritten as the integral of a well known series expansion

I probably mistranslated in French it’s said that way

[junior acorn]

what's the gist of the theorem?

[heady bluff]

Basically given certain conditions you can switch integral and sum (maybe beppo Levi’s theorem is more appropriate, idk)

[junior acorn]

aah

fubini

or rather a special case

i know what you mean, no worries

[heady bluff]

Why are you summing them ? What are you trying to achieve here ?

Are you trying to get the sum of x^(2k+1)/2k+1? The problème is that’s not going to work for x=1

[golden pilot]

the intervention

what

[heady bluff]

?

[vast citrus]

I can't see barely one

Many

😇😇

Bro even if you are not giving any hints

[heady bluff]

hint : $\frac{1}{2k+1}=\int_{0}^{1} x^{2k} dx$

[jovial nebula]

Rotor 😑

[heady bluff]

now you can exprès similarly 1/(2k+2) and 1/2k+3 for k>=0

[vast citrus]

x^2k+1/2k+1=1/2k+1

Integration 0 to 1 x^(2k+1)

[heady bluff]

using what I said the sum becomes $\frac{-1}{2} \sum_{k=0}^{+\infty} \left(\int_{0}^{1} x^{2k} -6x^{2k+1}+5x^{2k+2} dx \right)$

[jovial nebula]

Rotor 😑

[heady bluff]

@vast citrus

Now switch integral and sum (i won’t justify it here but it works, you can)

And think of the geometric series

(Here we are integrating on [0,1[ so |x|<1)

From there you will have an integral to calculate which will give you your result

You could’ve also just at the beginning rearranged the sim to get a telescopic sum which may have given easier calculations

$\sum_{k=0}^{+\infty} \frac{1}{2k+1}- \frac{6}{2k+2} +\frac{5}{2k+3}=\sum_{k=0}^{+\infty} 5\left(\frac{1}{2k+3} -\frac{1}{2k+1} \right) +6\left(\frac{1}{2k+1} -\frac{1}{2k+2} \right)$

[jovial nebula]

Rotor 😑

[heady bluff]

the first part is telescopic and for the second one you can do the integral trick

Sorry for yapping so much I hope this helps

@vast citrus

[heady bluff]

This method reminds me of a proof of the Basel problem

[vast citrus]

I see

I am reading and understanding

Slowly

Thanks rotor