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The sum can be represented in terms of the digamma function: S = -1/3Σ1/k - 1/(k+1/3) = -γ/3 - 1/3(-γ + Σ1/k - 1/(k+4/3-1)) = -1/3(ѱ(4/3) + γ) = -1/3(ѱ(1/3) + γ) - 1 By multiplication formula (z=1/3, k=3): ѱ(1/3) + ѱ(2/3) + ѱ(1) = 3(ѱ(1) - ln3) ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3) By reflection formula (z=1/3): ѱ(2/3) - ѱ(1/3) = πcot(π/3) = π/√3 So we have the linear relation: ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3) ѱ(2/3) - ѱ(1/3) = π/√3 ѱ(1/3) = -γ - 3ln(3)/2 - π/2√3 Hence S = ln(3)/2 + π/6√3 - 1 Alternatively you can use the Gauss digamma theorem
Nice solution. I would do the following: Define three series, A,B,C defined as the sums of respectively 1/(3n-1), 1/(3n), 1/(3n+1), each from n=1 to n=N. Then A+B+C=H_(3N)-1 and B=H_N/3 where H denotes the harmonic numbers. Now the series A-C is nice and can be easily computed to be 1-π/(3√3) using the cotangent residues trick from complex analysis. Then we can find the series we want, C-B, as ((A+B+C)-3B-(A-C))/2. From the approximation H_N≈ln N+γ we find lim H_(3N)-H_N=ln 3. So in total we get (ln 3-1-(1-π/(3√3)))/2 which agrees with your answer.
I've been working on an integral similar to this one but in the sum reverse the order of the terms and substitute a natural number k instead of 3. So far it doesn't seem to have a nice solution but I'll still try to genralise because I have nothing else to do with my life.
It's a great video ! The only thing I would like to see more is (even if it is simple and short) an explanation of why we can switch the sum and integral. It makes the video more satisfying for me, It kid of "legitimates" the result. thank you :)
Well done! 👌 (Maybe you would like to work a bit on your intonation as it sounds a little bit monotonous. But that may be just an interpretation by my own ears. ✌️)
@@harrydiv321 wow! Ok that's really interesting. I can't tell you for sure that's wrong but in youtube studio it is showing a like % of 98.9%. So it comes out to 183 likes to 2 dislikes. What do you make of it?
The sum can be represented in terms of the digamma function:
S = -1/3Σ1/k - 1/(k+1/3)
= -γ/3 - 1/3(-γ + Σ1/k - 1/(k+4/3-1))
= -1/3(ѱ(4/3) + γ)
= -1/3(ѱ(1/3) + γ) - 1
By multiplication formula (z=1/3, k=3):
ѱ(1/3) + ѱ(2/3) + ѱ(1) = 3(ѱ(1) - ln3)
ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3)
By reflection formula (z=1/3):
ѱ(2/3) - ѱ(1/3) = πcot(π/3) = π/√3
So we have the linear relation:
ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3)
ѱ(2/3) - ѱ(1/3) = π/√3
ѱ(1/3) = -γ - 3ln(3)/2 - π/2√3
Hence S = ln(3)/2 + π/6√3 - 1
Alternatively you can use the Gauss digamma theorem
Some serious info for my recollection right there! Excellent 👌
Nice solution. I would do the following: Define three series, A,B,C defined as the sums of respectively 1/(3n-1), 1/(3n), 1/(3n+1), each from n=1 to n=N. Then A+B+C=H_(3N)-1 and B=H_N/3 where H denotes the harmonic numbers. Now the series A-C is nice and can be easily computed to be 1-π/(3√3) using the cotangent residues trick from complex analysis. Then we can find the series we want, C-B, as ((A+B+C)-3B-(A-C))/2. From the approximation H_N≈ln N+γ we find lim H_(3N)-H_N=ln 3. So in total we get (ln 3-1-(1-π/(3√3)))/2 which agrees with your answer.
Nice 👍 thanks
I've been working on an integral similar to this one but in the sum reverse the order of the terms and substitute a natural number k instead of 3. So far it doesn't seem to have a nice solution but I'll still try to genralise because I have nothing else to do with my life.
It seems to be related to the Digamma function, so you might want to check that out.
It equals approximately -0.14839396162690884587003100526504 .
Brilliant approach
thanks!
It's a great video ! The only thing I would like to see more is (even if it is simple and short) an explanation of why we can switch the sum and integral. It makes the video more satisfying for me, It kid of "legitimates" the result. thank you :)
Thanks DJ! And it’s a good point too 👍
That’s the hardest part, the rest is easy af
@@owl3mathdo real maths bro, please man. It pains me to see someone that isn’t rigorous
Great job!
Hey Ron. Thanks! :)
That's fascinating ❤❤❤
Thanks!
This is easy af bro
Nice job!
Hey Mike. Thank you!
Well done! 👌
(Maybe you would like to work a bit on your intonation as it sounds a little bit monotonous. But that may be just an interpretation by my own ears. ✌️)
Thanks. And I think you’re right. I could use a better tone but it’s not my tendency
This question came in ny exam and i did cause i know the way
Nice :)
俺やったら出されても解けんな…
和分で解けた！
wtf is with the dislikes LOL
Hey Harry. Like in the comments? This one is doing well with views and thumbs up percentage is very high but the comment section hates it. 😆😂
@@owl3math i have the "return youtube dislikes" chrome plugin and this video has 356 dislikes for some reason
@@harrydiv321 wow! Ok that's really interesting. I can't tell you for sure that's wrong but in youtube studio it is showing a like % of 98.9%. So it comes out to 183 likes to 2 dislikes. What do you make of it?
@@owl3math ok that's weird, i guess its the extension
Ok good. That’s scary 🫣