Approximating $\sum_{k=1}^{\infty}\frac{\sin(k^{1/n})}{k}\text{ for }n\in\mathbb{N}$

Here is a more detailed reasoning: Using the Riemann-Stieltjes integral and the periodic Bernoulli polynomials $\tilde{B}_{n}(x)$, for $a > 0$

\begin{align*} \sum_{k=1}^{N} \frac{\sin(k^{1/n}a)}{k} &= \int_{1-}^{N} \frac{\sin(x^{1/n}a)}{x} \, d[x] \\ &= \int_{1}^{N} \frac{\sin(x^{1/n}a)}{x} \, dx - \int_{1-}^{N} \frac{\sin(x^{1/n}a)}{x} \, d\tilde{B}_{1}(x) \\ &= n \int_{1}^{a N^{1/n}} \frac{\sin x}{x} \, dx - \int_{1-}^{N^{1/n}} \frac{\sin(ax)}{x^{n}} \, d\tilde{B}_{1}(x^{n}). \end{align*}

So let us focus on the second term. Integrating by parts,

\begin{align*} - \int_{1-}^{N^{1/n}} \frac{\sin(ax)}{x^{n}} \, d\tilde{B}_{1}(x^{n}) &= \left[ -\frac{\sin(ax)}{x^{n}} \tilde{B}_{1}(x^{n}) \right]_{1-}^{N^{1/n}} + \int_{1}^{N^{1/n}} \left( \frac{\sin(ax)}{x^{n}} \right)' \tilde{B}_{1}(x^{n}) \, dx \\ &= \frac{\sin a}{2} + \int_{1}^{\infty} \frac{ax^{1/n}\cos(ax^{1/n}) - n \sin(ax^{1/n})}{n x^{2}} \tilde{B}_{1}(x) \, dx \\ &\qquad + \mathcal{O}(N^{-1+1/n}). \end{align*}

(But in fact, due the cancelling behavior, the error estimate may be improved further.) So when $n > 1$, combining together gives

$$ \sum_{k=1}^{N} \frac{\sin(a k^{1/n})}{k} = C + \mathcal{O}(\max \{ N^{-1+1/n}, N^{-1/n} \} ), $$

where

\begin{align*} C &= \sum_{k=1}^{\infty} \frac{\sin(a k^{1/n})}{k} \\ &= n \left( \frac{\pi}{2} - \mathrm{Si}(1) \right) + \frac{\sin a}{2} + \int_{1}^{\infty} \frac{ax^{1/n}\cos(ax^{1/n}) - n \sin(ax^{1/n})}{n x^{2}} \tilde{B}_{1}(x) \, dx. \end{align*}

And I highly suspect that the ever-growing error results from computation errors, not from the actual difference between the sum and the $N$-th partial sum.

Added. If you execute the following code,

  a = 1;
  n = 2;
  Quiet[N[Im[Sum[E^(I k^(1/n))/k, {k, 1, Infinity}]]]]
  Quiet[N[n (Pi/2 - SinIntegral[1]) + Sin[a]/2 + NIntegrate[(BernoulliB[1, FractionalPart[x]] (a x^(1/n) Cos[a x^(1/n)] - n Sin[a x^(1/n)]))/(n x^2), {x, 1, Infinity}]]]
  Quiet[N[n (Pi/2 - SinIntegral[1]) + Sin[a]/2 + NIntegrate[(BernoulliB[1, FractionalPart[x]] (a x^(1/n) Cos[a x^(1/n)] - n Sin[a x^(1/n)]))/(n x^2), {x, 1, Infinity}, WorkingPrecision -> 200, PrecisionGoal -> 50]]]
  Quiet[N[n (Pi/2 - SinIntegral[1]) + Sin[a]/2 + NIntegrate[(BernoulliB[1, FractionalPart[x]] (a x^(1/n) Cos[a x^(1/n)] - n Sin[a x^(1/n)]))/(n x^2), {x, 1, Infinity}, WorkingPrecision -> 400, PrecisionGoal -> 50]]]
  Quiet[N[n (Pi/2 - SinIntegral[1]) + Sin[a]/2 + NIntegrate[(BernoulliB[1, FractionalPart[x]] (a x^(1/n) Cos[a x^(1/n)] - n Sin[a x^(1/n)]))/(n x^2), {x, 1, Infinity}, WorkingPrecision -> 600, PrecisionGoal -> 50]]]
  Clear[a, n];

Then you will see how the numerical methods used in Mathematica are susceptible:

  1.71567
  1.72229
  1.71513
  1.71661
  1.71436

Consider the integral $$\int_1^k\frac{\sin x^{1/n}}xdx=n\int_1^{k^{1/n}}\frac{\sin t}t\ dt$$ by the change of variable $x=t^n$.

This is the Sine Integral.

It is closely related to your sum, and explains why the value of the exponent $n$ doesn't have such a big impact on the behavior.