Mathematics

This limit is asking you to recognize a derivative in disguise. Notice that the numerator $e^{1/n}-1$ and denominator $1/n$ have the same structure as the definition of a derivative: $\lim_{h\to 0}\frac{f(h)-f(0)}{h}$ where $f(x)=e^x$. As $n\to\infty$, the quantity $1/n\to 0$, so you're really evaluating the derivative of $e^x$ at $x=0$. Once you find this limit, use the divergence test to determine whether the series $\sum a_n$ can possibly converge.

This limit is asking you to recognize a derivative in disguise. Notice that the numerator $e^{1/n}-1$ and denominator $1/n$ have the same structure as the definition of a derivative: $\lim_{h\to 0}\frac{f(h)-f(0)}{h}$ where $f(x)=e^x$. As $n\to\infty$, the quantity $1/n\to 0$, so you're really evaluating the derivative of $e^x$ at $x=0$. Once you find this limit, use the divergence test to determine whether the series $\sum a_n$ can possibly converge.