Clearly, $a>1$ .

Also, note that : $\dfrac{1}{2} > \dfrac{1}{3} > \dfrac{1}{4} > ...$

Hence :

$a < 1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...$

$a < 2-\dfrac{1}{2}$

So, $1<a<1.5 \implies [a] = 1$ .

Relevant wiki: Riemann Zeta Function

We note that:

$\begin{aligned} 1 < \sum_{n=1}^\infty \frac 1{n^n} & < \sum_{n=1}^\infty \frac 1{n^2} \\ \implies 1 < a & < {\color{#3D99F6}\zeta (2)} = \frac {\pi^2}6 \approx 1.645 & \small \color{#3D99F6} \text{where }\zeta(\cdot) \text{ denotes the Riemann zeta function.} \end{aligned}$

$\implies \lfloor a\rfloor = \boxed{1}$

Accurately we have,

$\boxed{\boxed{1.29129}}$

From this cpp code

include<iostream>

include<math.h>

using namespace std;

int main()

{

float n,sum;

int i,m;


cout<<"Enter the no. of terms";

cin>>n;

for(i=1;i<=n;i++)

{

    sum=sum+1.0/(pow(i,i));

}

m=sum;

cout<<sum;

return 0;

}