Can we directly use, a sub n < or = |a sub n| and then sum both sides from n=1 to infinity. As RHS is given as convergent, so LHS (the required sum) is convergent..??
No, since the intuition behind the comparison test only applies to series of non-negative terms, and a_n could be negative, hence your argument does not hold.
Thank you, clean and beautiful!
thank you so much that was so profound and on point
i dont understand how you magically find a negative -|an| to remove it from the other series
Can we directly use, a sub n < or = |a sub n| and then sum both sides from n=1 to infinity. As RHS is given as convergent, so LHS (the required sum) is convergent..??
No, since the intuition behind the comparison test only applies to series of non-negative terms, and a_n could be negative, hence your argument does not hold.
Thank you.
Why do you assume that an is monotone growing?
No such assumption was made. :-)
@@slcmathpc Sorry I didn't understand it.👍
Excellent
Thank you
Very nice sir
Just had to be number 1😋