P-Series - Proof

[@prasannakumarz]

Superrr sir.my maths sir is nothing infront of ur explanation

[@shubham-tr8ep]

Good good

[@tangpiseth8416]

what if P is less than 0? then the series diverges right?

[@AryanGowda-s7m]

yes u can use test for divergence and check

[@dannyboy900102]

ah, that makes so much sense. Thanks for the helpful video!!

[@mahmoudmaher4201]

thank you sir for the clear explanation, i think a got the idea now, but just a quick question, in the case of the convergence of the series, does the integral test really tell us at what value the series will converge to at infinity, or the integral test only tell us that series will converge, for example the series of 1/n^2. if we integrate that from 1 to infinity will get the result to be 1, so does that mean the sequence 1/n^2 will converge to 1 also, or is this just telling us that it does converge, but we don't know what it will converge to?

[@slcmathpc]

If you look at the integral test geometrically, you can see that the improper integral gives an approximation to the exact value of the corresponding infinite series, so in the case that you mentioned, the integral test not only tells you that the series of 1/n^2 converges, but it tells you that its value is approximately equal to 1; in fact, the exact value of this infinite series is (π^2)/6, which is approximately 1.645. As you can see, the approximate value of 1 given by the integral test is fairly close to the exact value of the series. Oh and also, if you just sum the first two terms of the series, you obtain 1+1/4=1.25, and since the series contains only strictly positive terms, you can see right away that its exact value is obviously greater than 1.25. :-)

[@mahmoudmaher4201]

@@slcmathpc thank you sir for the clear explanation, but I was thinking about the graphical representation of the improper integral to just show us that the area under the curve of the function can be indicator to see if the series converges or not, but not to give us the approximation of the series, how are they related, if we think of series the same way that you have explained earlier to be the sum of discrete points on a continuous function, then how can we directly apply that to get the actual value of the series, how could be the area under the curve of a continuous function f(x) is related to summing y values of that function on discrete moments, this is really confusing

[@slcmathpc]

The integral test only offers an approximation; it cannot offer the exact value of the series. To find the exact value of the series, you need some new tricks! ;-)

[@mahmoudmaher4201]

@@slcmathpc thank you sir for the clear information and for that great material

[@빈츠-l9h]

Awesome!

[@GorgeousFortress]

Very nice video, I find it it difficult to find concise and accurate proofs for advance calculus in video form.

[@princemusonda2350]

Great👍

[@edmondscott7444]

Sir. You stated if p = 1 dgt, but in my opinion did not prove it. However your use of the integral test is very clear.

[@slcmathpc]

I proved the result in an earlier video; it is a direct application of the integral test.

[@boxxer221]

Great explanation.

[@navyakakarlapudi4505]

Thank you very much sir . Nice explanation..

[@deeputhattil5149]

Thanks for your help in p series sir.

[@Ishtiyaq_Mir]

Great

[@uttpalkant2888]

Thanks it is very useful

[@khushiraj8040]

Thanku sir

[@austinchen6024]

Almost as bad as T-Series.