I want to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^{\phi(n)}}$.
where $\phi$ is Euler's totient.
So for example $\frac{1}{1^1} + \frac{1}{2^1} + \frac{1}{3^2} +\frac{1}{4^2} + \frac{1}{5^4} + \frac{1}{6^2}$ etc...
I also want to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^{\lambda(n)}}$.
where $\lambda(n)$ is Carmichael's Lambda Function
I don't have a nice calculator so I can't approximate the values properly, but ideally I want to find a closed form.
I think it would end up being something like $\frac{1}{4} + y\frac{\pi^x}{k}$ where $y,x$and $k$ are different real numbers. The $\frac{1}{4}$ is by splitting the $\frac{1}{2}$ into $2 \times \frac{1}{2^2}$ where one gets used in the series and the other goes outside.
The reason I believe there is a closed form is because all the even Basel problems have closed answers $\displaystyle\sum_{n=1}^\infty \frac{1}{n^{2x}}$ and the totient and lambda functions also produce even numbers.