8:21 I think it is n=m+1 and Because if n=m the absolute value is zero and it not give us that consequence. But if we apply n=m+1 we will get that consequence, that the series (summation(xn)) converges iff the sequence (xn) converges to 0
Sir, for the convergence of S_n = \sum{1/n^2}, can we think of it as S_n \leq < 1/[(n)(n-1)] = t_n \forall n>1, where t_n converges, hence S_n also converges for all n \in N ?
At 8:29, I think it should be "suppose m=(n-1).
22:46 that girl with glasses can't get her eyes off the camera man!🤣🤣
Damn it
Thank You sir, Your lectures are so useful.
8:21 I think it is n=m+1 and Because if n=m the absolute value is zero and it not give us that consequence. But if we apply n=m+1 we will get that consequence, that the series (summation(xn)) converges iff the sequence (xn) converges to 0
Is it meant for bsc students as well?
Sir, for the convergence of S_n = \sum{1/n^2}, can we think of it as S_n \leq < 1/[(n)(n-1)] = t_n \forall n>1, where t_n converges, hence S_n also converges for all n \in N ?
Sir
If n= m then how Sn-Sm is xn??
It simply follows from the fact that xn=Sn-Sn-1
May you explain what is, there exists £>0 and some other conditions like n>=m. where it come from.
He already has in previous lectures
Look for the definition of Cauchy Sequences.
yeah bro start from lec8
nice sir easy understandable thank u sir
May you explain what is there exists £>0 and some other conditions . where it come from.
@@BaoBeans refer from lecture 7