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Happy🥰Teachers'❤️Day Sir!🥳 Your guidance😇 and support 🤞🏻have shaped the future of many students✨, and we are all grateful 🙏🏻for your dedication😌. _May your day be as special😍 as the impact you've had on your students😌._ Thank you 🫶🏻for being such an incredible teacher❤️.
@@LUc942 Find the expression u(n) and u(n+1) then find lim n tends to infinity u(n) and then find lim n tends to infinity u(n+1)/u(n). Then U can solve it.
Sir I'm in second semester and I watch your video regularly based on mathematics.I want to tell you one thing, should I start preparing the jam from now?
B option is correct because after putting x=1 , Alternating series will be { 1 -1/3+ 1/5- 1/7....................1/(2n-1) } And look at this series it follows Leibnitz theorem i.e 1/(2n-1) >= 0 1/(2n+1)
B option is correct because after putting x=1 , Alternating series will be { 1 -1/3+ 1/5- 1/7....................1/(2n-1) } And look at this series it follows Leibnitz theorem i.e 1/(2n-1) >= 0 1/(2n+1)
SIR HOW LOG N /N SQUARE IS DECREASING SEQUENCE BECAUSE AFTER CALCULATING IT'S MAXIMUM AND MINIMUM VALUE WE CAN'T SAY IT'S DECREASING OR INCREASING. KINDLY CLEAR IT SIR
B option is correct because after putting x=1 , Alternating series will be { 1 -1/3+ 1/5- 1/7....................1/(2n-1) } And look at this series it follows Leibnitz theorem i.e 1/(2n-1) >= 0 1/(2n+1)
Thank You for Watching the Video !
Get CSIR NET, IIT JAM, GATE, RPSC Courses, Test Series, Video Lectures, etc on our Mathscare App
🔹Android - play.google.com/store/apps/details?id=com.mathscare.app
🔹IOS - apps.apple.com/in/app/mathscare/id6467857729
🔹Telegram - telegram.me/mathsbygpsir
Thanks
Watching from NSUT
+1 bhai
greetings from PEC 1st year
Option B ✅
Watching from pec chandigarh batch 2024-2028
@@kalashdeep6623 myself from Punjab University lmao
@@divyanka4927me too great coincidence 😅
sher
@@divyanka4927important batado kya aana h ? 12 ko maths wale m
Greetings from UIET chandigarh
Same bro
Same bro
Thankyou so much sir 🌺, Correct answer is option b 😊
sir 13:08 how did u wrote 1/n^n-1 (n tending to infinity) 0 despite (infinity)^infinity
Option 1 》|X|
did you get x^2 in the end of D'Alemberts test?
@@is_math_hard yes
Happy🥰Teachers'❤️Day Sir!🥳
Your guidance😇 and support 🤞🏻have shaped the future of many students✨, and we are all grateful 🙏🏻for your dedication😌.
_May your day be as special😍 as the impact you've had on your students😌._
Thank you 🫶🏻for being such an incredible teacher❤️.
Hello Rupam
@AbhijeetMina-f6o Byeee
@@rupamthakur2002 kyu , esa mat karo
@@rupamthakur2002 your dp looking sexy
Option b correct
Thankyou sir
Ksise
Answer will be option (B).
Kaise
@@LUc942 Find the expression u(n) and u(n+1) then find lim n tends to infinity u(n) and then find lim n tends to infinity u(n+1)/u(n). Then U can solve it.
@@priyansumohanty9303 bro bro ye toh x aa rha h💀
@@priyansumohanty9303 axa axa dhanyawad aapka🙏🙏🙏🙏
@@priyansumohanty9303 bro answer B nhi A hoga kiyoki x≤1 mai x=1 bhi aa rha h jisme ratio test fail ho jata h
Isiliye x
Good job 👌
Sir I'm in second semester and I watch your video regularly based on mathematics.I want to tell you one thing, should I start preparing the jam from now?
Thank you Sir 🙏
Sir option A is correct in last given question
B option is correct because after putting x=1 , Alternating series will be { 1 -1/3+ 1/5- 1/7....................1/(2n-1) } And look at this series it follows Leibnitz theorem i.e 1/(2n-1) >= 0
1/(2n+1)
Thanks 🙏
i am from pcce college in goa thank you sir
Sir, please put English captions..
It will be helpful for those students who don't know Hindi..
Sir !!
Kummer's test [ INFINIT SERiEs ] par videos...
🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Option a is correct
sir apka content bhot achha h pr please bolne ki speed halki slow rkhiye ap bhot zada fast bolte h. Thanks for all the work you are doing 😋
Bhai mai toh fir bhi 2x pr dekta hu😅
@@deveshsarode Because you are a legend
Me as well watching at. 2X 😂😅
10 SEC
first, use Leibnitz for Convergence,
second, use ratio test to find x values
Option A is correct ❄️❄️
B galat h ek baar dobara ratio test pd lo😂
x^2
M bhi toh socho y aa kse rha h b option 😂
B option is correct because after putting x=1 , Alternating series will be { 1 -1/3+ 1/5- 1/7....................1/(2n-1) } And look at this series it follows Leibnitz theorem i.e 1/(2n-1) >= 0
1/(2n+1)
Ratio test is working on only positive series but this is alternative series bro , update your concepts clarity. 😅
bro in each and every ratio test terms were positive, this is an alternating series
Hallo sir please riply me ye video de hmm GTU University hai isame kam lgela aapka.. sare video..
PARABOLOID bata dijiye sir 🙏
Sir 🙏🙏
Please SC Mallick and Savita Arora se pdhaiye. ( BHU UNIVERSITY)
SIR HOW LOG N /N SQUARE IS DECREASING SEQUENCE BECAUSE AFTER CALCULATING IT'S MAXIMUM AND MINIMUM VALUE WE CAN'T SAY IT'S DECREASING OR INCREASING.
KINDLY CLEAR IT SIR
Option B ..tq sir ji
Kaise?
@@AkashSingh-qs3ek aapka un kitna aara h
Sorry 1st option is correct✅✔ answer h
Qki x=1 pr divergent aa rha h by second logarithium test
D alembert ratio test se x^20 to ye x=1 pr divergent hoga to ye series only x
@@AkashSingh-qs3ek hn a is correct
The answer is: Bombay (B)
how did you guys solve the last question?
I think the option is "A" coz Un should be less than one
Option b ~ how?
B option is correct because after putting x=1 , Alternating series will be { 1 -1/3+ 1/5- 1/7....................1/(2n-1) } And look at this series it follows Leibnitz theorem i.e 1/(2n-1) >= 0
1/(2n+1)
B
☑️
B
Kaise
Option B
Kaise?
Sir samjh nhi aa raha
Sir aap zabardasti karke bas answer la rahe hai jab limit put karte hai n me to uper koi value nahi honi chahiye n ke terms me. Thank you
option A
Option a
|x|
Option b
option d
Thanks you sir from cgc landran
First
b
A
Option B
B
A
B
B
B
B