determine if $\sum^{\infty}_{n=1} \frac{n}{e^n}$ converges determine if $\sum^{\infty}_{n=1} \frac{n}{e^n}$ converges. I used the ratio test and eventually came to that: $\frac{n+1}{en}$ which is approximatley equal to $\frac{1}{e}$ which is less than 1. Therefore it converges
Is this correct?
 A: Yes, your method is correct. One can also find the exact sum quite easily in various ways. Here's one example through differentiation:
$$\sum_{n = 1}^{\infty} x^n = \frac{x}{1 - x}  = \frac{1}{1-x} - 1 \qquad \text{for $|x| < 1$}$$
$$ \sum_{n = 1}^{\infty} nx^{n-1} = \frac{1}{(1 - x)^2} \qquad \text{differentiating both sides}$$
$$ \sum_{n = 1}^{\infty} nx^n = \frac{x}{(1 - x)^2}$$
Thus, when $x = \frac{1}{e}$, we get the sum to be $\frac{\frac{1}{e}}{(1 - \frac1e)^2} = \frac{e}{(e - 1)^2}$
Alternatively without calculus:
Notice that $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{e^n} = \displaystyle \sum_{k = 1}^{\infty} S_k$, where $S_k = \displaystyle \sum_{n = k}^{\infty} \frac{1}{e^n}$
Then $S_k = \frac{1/e^k}{1 - 1/e}$ by geometric series formula. Again by geometric series formula, $\displaystyle \sum_{k = 1}^{\infty} S_k = \frac{\frac{1/e}{1 - 1/e}}{1 - 1/e} = \frac{1/e}{(1 - 1/e)^2} = \frac{e}{(e - 1)^2}$
A: Michael T. has posted the usual argument involving differentiation.  Here a probabilistic argument.
When you throw a die, you have a $1/6$ chance of getting an ace.  So average number of throws to get an ace is $6$.  Therefore
$$
6 = \sum_{n=1}^\infty n\cdot \Pr(\text{number of throws}=n) = \sum_{n=1}^\infty n\cdot\left(\frac56\right)^{n-1} \cdot\frac 1 6.
$$
If the probability of failure on each trial were $1/e$ instead of $5/6$, then the probability of success on each trial would be $1-(1/e)=(e-1)/e$ and the average number of trials to get a success would be the reciprocal of that: $e/(e-1)$.
So put
\begin{align}
1/e & \text{in place of }5/6, \\
(e-1)/e & \text{in place of }1/6, \\
e/(e-1) & \text{in place of }6.
\end{align}
We get
$$
\frac{e}{e-1} = \sum_{n=1}^\infty n\left(\frac 1 e\right)^{n-1}\cdot\frac{e-1}{e}.
$$
Divide both sides by $e-1$:
$$
\frac{e}{(e-1)^2}= \sum_{n=1}^\infty \frac{n}{e^n}.
$$
A: Another approach which, perhaps, can be easier to apply in this particular case, because of the exponential: the $\;n$-th root test:
$$\sqrt[n]{\frac n{e^n}}=\frac{\sqrt[n] n}e\xrightarrow[n\to\infty]{}\frac1e<1\implies\;\text{the series converges}$$
