Analyzing the Series $ \sum_{k=1}^{+\infty} \sqrt[k]{3} $

We consider the infinite series

$$ \sum_{k=1}^{+\infty} \sqrt[k]{3} $$

To begin, we examine the general term:

$$ a_k = \sqrt[k]{3} = 3^{1/k} $$

This expression can be rewritten as:

$$ a_k = 3^{1/k} $$

We now study the behavior of the general term as \( k \to \infty \):

$$ \lim_{k \to \infty} 3^{1/k} $$

To evaluate this limit, we express it in logarithmic form:

$$ \lim_{k \to \infty} e^{\ln(3^{1/k})} $$

Applying the logarithmic identity \( \ln(a^b) = b \ln a \), we get:

$$ \lim_{k \to \infty} e^{\frac{1}{k} \ln 3} $$

Since \( \ln 3 \) is a constant and \( \frac{1}{k} \to 0 \) as \( k \to \infty \), the exponent tends to zero. Hence,

$$ \lim_{k \to \infty} e^{\frac{1}{k} \ln 3} = e^0 = 1 $$

This tells us that the general term does not approach zero:

$$ \lim_{k \to \infty} a_k = 1 $$

This is a crucial observation. According to the convergence criterion, a necessary (though not sufficient) condition for the convergence of an infinite series \( \sum a_k \) is that the general term tends to zero:

$$ \lim_{k \to \infty} a_k = 0 $$

Since this condition is not met, we conclude that the series diverges:

$$ \lim_{k \to \infty} \sqrt[k]{3} = 1 \ne 0 $$

Therefore, the series is divergent:

$$ \sum_{k=1}^{+\infty} \sqrt[k]{3} \quad \text{diverges} $$

That completes the analysis.