given equation is convergent if k>1. $$ \sum_{n=1}^\infty \frac{1}{n^k} $$.
can somebody tell me that how can i prove?
My intuition:
For k=1 $$\log(1+x) = x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 ....$$ if we substitute x = -1.then, $$1+\frac{1}2+\frac{1}3 ... = -\log(0) = \infty$$ it is very hard to digest that equation converges even if k = 1.000000001
Cauchy condensation test is a very popular tool for this problem. The theorem states:
$$\text{For a non negative, non increasing series } a_n \text{ :the sum } \sum\limits_{n=1}^\infty a_n \text{ converges iff the sum } \sum\limits_{n=1}^\infty 2^na_{2^n} \text{ converges.}$$
Using this THM, assuming the knowledge of the conditions for convergence of a geometric series - the result follows.
