Let's consider series from n=1 to inf (x^(2n+1)/2^(2n-1)). This is geometric series with sum equal to 2x^3/(4-x^2). Now take derivative both sides and put x=1. You will get that the sum you are looking for is (2x^3/(4-x^2))' at x=1 which is 22/9
Non-terminating decimal...unlike pi non-terminating non repeating decimal. Strange things happening out there in infinity....maintain your point of conciousness and kick these paradoxes outtta the way! I don't know, I'm nuts...,but I love ya.
Respected Sir, your videos are very informative. 👍 We get a lot to learn. 😇 I recently got to know about RAMANUJAN'S SUMMATION paradox, where 1+2+3+4+5+... = -1/12. It would be helpful, if you could share yor views on it. 😇 Advance thanks. 👍
You are very welcome. So nice of you. Thank you for your feedback! Cheers! You are awesome Leo 😀 Keep it up 👍 I've already uploaded the video. Please check out the link: ruclips.net/video/pgxbm6Vjhkk/видео.html
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Sir please check the previous video on quadrilateral. I commented there by solving it by another method please check it and comment on that is it correct or not
...rilevo un errore... non si puó "trattare" la somma infinita come se si sapesse giá che converge. nel video questa cosa non si dice da nessuna parte... perció, prima bisogna dimostrare che la serie converge ad S e poi fare le operazioni. Senza questa prova, non puó eseguire brutalmente la differenza tra le due scritture . La S significa somma finita, ma tu non hai dimostrato che la serie é convergente!!!!!! quindi é un azzardo fare somme o differenze con infiniti termini senza avere la certezza che si potra parlare di Somma infinita e non di somma infinita.👋👋👋
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Let's consider series from n=1 to inf (x^(2n+1)/2^(2n-1)). This is geometric series with sum equal to 2x^3/(4-x^2). Now take derivative both sides and put x=1. You will get that the sum you are looking for is (2x^3/(4-x^2))' at x=1 which is 22/9
this problem can be divided into infinite number of G-series with different first terms and 1/4 as common ratio.
S1 = 3 [1/2+1/8+1/32+1/128+1/512+..........] = 3[(4/3)(1/2)] = 2 using Sg = a0/(1-r)
S2 = 2[1/8+1/32+1/128+1/512+..................] = 2[(4/3)(1/8)]
S3 = 2[1/32+1/128+1/512+...........................] = 2[(4/3)(1/32)]
S4 = 2[1/128+1/512+.....................................] = 2[(4/3)(1/128)]
...
....
Let S = S2+S3+S4+........ = 2[(4/3) {1/8+1/32+1/512+.........}] = 2[(4/3){(4/3)(1/8)}] = 4/9
Total sum of given G-series T = S1 + S = 2 + 4/9 = 22/9
..............S1.................
1/2 + 1/2 + 1/2 .........S2........
1/8 + 1/8 + 1/8 + 1/8 + 1/8 ...........S3........
1/32 + 1/32 +1/32 + 1/32 + 1/32 + 1/32 + 1/32 ...........S4.......
1/128 + 1/128 +1/128 + 1/128 + 1/128 + 1/128 + 1/128 + 1/128 + 1/128 ...........S5.......
1/512 + 1/512 +1/512 + 1/512 + 1/512 + 1/512 + 1/512 + 1/512 + 1/512 +1/512 +1/512
....
....
....
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Ans around 11/4
Non-terminating decimal...unlike pi non-terminating non repeating decimal. Strange things happening out there in infinity....maintain your point of conciousness and kick these paradoxes outtta the way! I don't know, I'm nuts...,but I love ya.
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Respected Sir, your videos are very informative. 👍 We get a lot to learn. 😇
I recently got to know about RAMANUJAN'S SUMMATION paradox, where 1+2+3+4+5+... = -1/12.
It would be helpful, if you could share yor views on it. 😇 Advance thanks. 👍
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I've already uploaded the video. Please check out the link:
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@@PreMath 👍👍 Thank you very much, Sir. 👍👍 I was always thinking whether I should go ahead and do my Masters in Mathematics. Your videos are so motivational. 👍👍 Thank you once again. 😇👍
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Nice question, today I learn again one thing
I always found myself weak in series and log related questions
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very well explained, thanks for sharing, 242 math here
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Sir please check the previous video on quadrilateral. I commented there by solving it by another method please check it and comment on that is it correct or not
Please see the reply for that.
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This question is from what chapter and what class ??
Power series. S-rS type.
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...rilevo un errore...
non si puó "trattare" la somma infinita come se si sapesse giá che converge.
nel video questa cosa non si dice da nessuna parte...
perció, prima bisogna dimostrare che la serie converge ad S e poi fare le operazioni.
Senza questa prova, non puó eseguire brutalmente la differenza tra le due scritture .
La S significa somma finita, ma tu non hai dimostrato che la serie é convergente!!!!!! quindi é un azzardo fare somme o differenze con infiniti termini senza avere la certezza che si potra parlare di Somma infinita e non di somma infinita.👋👋👋
Interrested
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Expn.=309/108ans
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22/9,diventa la somma di una serie geometrica e di una serie geometrica derivata....OK RISPOSTA GIUSTA!!!!
Ottimo lavoro Giuseppe!
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Jaake Sachin sir lecture no 8 dekh lo halwa lagenge aise sawaal🔥
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Done
parfait
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wow
22÷9=2.4444444444444444444444.