1. The problem is to find the Moment Generating Function (MGF) for a random variable $X$ with probabilities $P(X=k) = \frac{1}{k}$ for $k=1,2,3,\ldots$ using the geometric series. 2. First, we note that the probabilities $P(X=k)=\frac{1}{k}$ for $k=1,2,3,\ldots$ do not sum to 1, so this cannot be a valid probability distribution. However, if you meant the geometric distribution which has the PMF $P(X=k) = p(1-p)^{k-1}$ for $k=1,2,3,\ldots$, then we can proceed. 3. The MGF of a geometric random variable with parameter $p$ is defined as: $$ M_X(t) = E(e^{tX}) = \sum_{k=1}^\infty e^{tk} p(1-p)^{k-1} $$ 4. Factor $p$ outside the summation: $$ M_X(t) = p \sum_{k=1}^\infty \left(e^{t}(1-p)\right)^{k-1} e^{t} $$ 5. Rewrite the summation as: $$ M_X(t) = p e^{t} \sum_{k=0}^\infty \left(e^{t}(1-p)\right)^k $$ 6. Since this is a geometric series with ratio $r = e^{t}(1-p)$, which converges for $|r| < 1$, the sum is: $$ \sum_{k=0}^\infty r^k = \frac{1}{1-r} $$ 7. Substitute back: $$ M_X(t) = \frac{p e^{t}}{1 - e^{t}(1-p)} $$ 8. Thus, the MGF of the geometric distribution with parameter $p$ is: $$ \boxed{M_X(t) = \frac{p e^{t}}{1 - (1-p) e^{t}}} $$ 9. If the question was instead about $P(X=k) = \frac{1}{k}$, then it is not a probability distribution and the MGF doesn't exist in standard form.