I take it that you are dealing with the exponential distribution with $ f_X(x) = \lambda \mathrm{e}^{-\lambda x} [x > 0] $ Assuming all elements of the sample $\{x_1,x_2,\ldots,x_n\}$ are positive, the log-likelihood reads: $ n \log \lambda - \lambda \sum_{k=1}^n x_k $ which attains its maximum exactly at $\lambda = \frac{n}{\sum_{k=1}^n x_k}$.
Now to computation of the expectation of the MLE: $\begin{eqnarray} \mathbb{E}\left(\frac{n}{X_1+X_2+\cdots+X_n}\right) &=& n \mathbb{E}\left(\frac{1}{X_1+X_2+\cdots+X_n}\right) \\ &=& n \mathbb{E}\left( \int_0^\infty \exp\left(-t(X_1+\cdots+X_n)\right) \mathrm{d}t \right) \\ &=& n \int_0^\infty \mathbb{E}\left( \exp\left(-t(X_1+\cdots+X_n)\right) \right) \mathrm{d}t \\ &\stackrel{\text{indep.}}{=}& n \int_0^\infty \left(\mathbb{E}\left( \exp\left(-t X_1\right) \right)\right)^n \mathrm{d}t \\ &=& n \int_0^\infty \left(\frac{\lambda}{t+\lambda}\right)^n \mathrm{d}t = \left. -\frac{n}{n-1} \frac{\lambda^n}{(t+\lambda)^{n-1}} \right|_{0}^\infty \\ &=& \frac{n}{n-1} \lambda \end{eqnarray} $ Thus the MLE is biased.