2021/043) Find the largest integer n for which $\sum_{k=1}^\infty \frac{1}{k^n}$ diverges

Ans n = 1.

To prove that same we show that if n = 1 then this diverges and if n =2 this converges

To prove that it diverges for n = 1

Let the sum be S

We have $S = \sum_{k=1}^\infty \frac{1}{k^n}$

Or 

$S = 1 +  \sum_{k=1}^\infty (\sum_{i=2^{k-1}+1}^{2^{k}} \frac{1}{i^n})$

 $S = 1 + \sum_{k=1}^\infty s_k\cdots(1)$

where $s_k = \sum_{i=2^{k-1} +1}^{2^{k}} \frac{1}{i^n}$

Basically we have broken the sum to smaller groups with each group $k^{th}$ group containing the sum of reciprocal of $2^{k-1} +1$ to $2^{k}$ we have separated 1 and 1/2 so that k can start from 1 

Now let us compute $s_k = \sum_{i=2^{k-1} +1}^{2^{k}} \frac{1}{i^n}$

The above has got $2^{k-1}$ terms 

and for $i < 2^k$ we have $ \frac{1}{i} > \frac{ 1}{2^k}$

Hence $s_k >= 2^{k-1} * \frac{ 1}{2^k}$

or $s_k >= \frac{ 1}{2}$

Putting in (1) we get

$S = 1 + \sum_{k=1}^\infty s_k = 1 + \sum_{k=1}^\infty \frac{1}{2} $

Hence it diverges 

Now let us prove that it converges for n =2

For $n >=2$

We have $n^2 > n(n-1)$ or $\frac{1}{n^2} < \frac{1}{n(n-1)}$

We have $n^2 > n(n-1)$ or $\frac{1}{n^2} < \frac{1}{n(n-1)}$

as $\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$

so $\frac{1}{n^2} <  \frac{1}{n-1} - \frac{1}{n}$

So $\sum_{k=1}^\infty \frac{1}{k^n}$

$= 1 + \sum_{k=2}^\infty \frac{1}{k^n}$

$<  1 + \sum_{k=2}^\infty(\frac{1}{k-1} - \frac{1}{k}$

$<= 1 + (1- \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3} ) + \cdots$

$ < 2$